In order not to leave this question unanswered, let me prove the claim along
the lines I've suggested in the comments.
Let us agree on a few notations:
Let $n$ and $m$ be two nonnegative integers. Let $A=\left( a_{i,j}\right)
_{1\leq i\leq n,\ 1\leq j\leq m}$ be an $n\times m$-matrix (over some ring).
Let $U=\left\{ u_{1}<u_{2}<\cdots<u_{p}\right\} $ be a subset of $\left\{
1,2,\ldots,n\right\} $, and let $V=\left\{ v_{1}<v_{2}<\cdots<v_{q}\right\}
$ be a subset of $\left\{ 1,2,\ldots,m\right\} $. Then, $A_{U,V}$ shall
denote the submatrix $\left( a_{u_{i},v_{j}}\right) _{1\leq i\leq p,\ 1\leq
j\leq q}$ of $A$. (This is the matrix obtained from $A$ by crossing out all
rows except for the rows numbered $u_{1},u_{2},\ldots,u_{p}$ and crossing out
all columns except for the columns numbered $v_{1},v_{2},\ldots,v_{q}$.) For
example,
\begin{equation}
\begin{pmatrix}
a_{1} & a_{2} & a_{3} & a_{4}\\
b_{1} & b_{2} & b_{3} & b_{4}\\
c_{1} & c_{2} & c_{3} & c_{4}\\
d_{1} & d_{2} & d_{3} & d_{4}
\end{pmatrix}
_{\left\{ 1,3,4\right\} ,\left\{ 2,4\right\} }
=
\begin{pmatrix}
a_{2} & a_{4}\\
c_{2} & c_{4}\\
d_{2} & d_{4}
\end{pmatrix} .
\end{equation}
If $n$ is a nonnegative integer, then $I_n$ will denote the $n\times n$
identity matrix (over whatever ring we are working in).
Fix a nonnegative integer $n$ and a field $\mathbb{F}$.
We shall use the following known fact:
Theorem 1. Let $\mathbb{K}$ be a commutative ring. Let $A$ be an $n\times
n$-matrix over $\mathbb{K}$. Let $x\in\mathbb{K}$. Then,
\begin{align}
\det\left( A+xI_n \right) & =\sum_{P\subseteq\left\{ 1,2,\ldots
,n\right\} }\det\left( A_{P,P}\right) x^{n-\left\vert P\right\vert
}
\label{darij.eq.t1.1}
\tag{1}
\\
& =\sum_{k=0}^{n}\left( \sum_{\substack{P\subseteq\left\{ 1,2,\ldots
,n\right\} ;\\\left\vert P\right\vert =n-k}}\det\left( A_{P,P}\right)
\right) x^{k}.
\label{darij.eq.t1.2}
\tag{2}
\end{align}
Theorem 1 appears, e.g., as Corollary 6.164 in my Notes on the combinatorial
fundamentals of algebra, in the version of 10th January
2019 (where I
use the more cumbersome notation $\operatorname*{sub}\nolimits_{w\left(
P\right) }^{w\left( P\right) }A$ instead of $A_{P,P}$). $\blacksquare$
Corollary 2. Let $A$ be an $n\times n$-matrix over a field $\mathbb{F}$.
Let $r\in\left\{ 0,1,\ldots,n\right\} $. Consider the $n\times n$-matrix
$tI_n +A$ over the polynomial ring $\mathbb{F}\left[ t\right] $. Its
determinant $\det\left( tI_n +A\right) $ is a polynomial in $\mathbb{F}
\left[ t\right] $. Then,
\begin{align}
& \left( \text{the sum of all principal }r\times r\text{-minors of }A\right)
\nonumber\\
& =\left( \text{the coefficient of }t^{n-r}\text{ in the polynomial }
\det\left( tI_n +A\right) \right) .
\end{align}
Proof of Corollary 2. We have $r\in\left\{ 0,1,\ldots,n\right\} $, thus
$n-r\in\left\{ 0,1,\ldots,n\right\} $. Also, from $tI_n +A=A+tI_n $, we
obtain
\begin{equation}
\det\left( tI_n +A\right) =\det\left( A+tI_n \right) =\sum_{k=0}
^{n}\left( \sum_{\substack{P\subseteq\left\{ 1,2,\ldots,n\right\}
;\\\left\vert P\right\vert =n-k}}\det\left( A_{P,P}\right) \right) t^{k}
\end{equation}
(by \eqref{darij.eq.t1.2}, applied to $\mathbb{K}=\mathbb{F}\left[ t\right]
$ and $x=t$). Hence, for each $k\in\left\{ 0,1,\ldots,n\right\} $, we have
\begin{align*}
& \left( \text{the coefficient of }t^{k}\text{ in the polynomial }
\det\left( tI_n +A\right) \right) \\
& =\sum_{\substack{P\subseteq\left\{ 1,2,\ldots,n\right\} ;\\\left\vert
P\right\vert =n-k}}\det\left( A_{P,P}\right) .
\end{align*}
We can apply this to $k=n-r$ (since $n-r\in\left\{ 0,1,\ldots,n\right\} $)
and thus obtain
\begin{align*}
& \left( \text{the coefficient of }t^{n-r}\text{ in the polynomial }
\det\left( tI_n +A\right) \right) \\
& =\sum_{\substack{P\subseteq\left\{ 1,2,\ldots,n\right\} ;\\\left\vert
P\right\vert =n-\left( n-r\right) }}\det\left( A_{P,P}\right)
=\sum_{\substack{P\subseteq\left\{ 1,2,\ldots,n\right\} ;\\\left\vert
P\right\vert =r}}\det\left( A_{P,P}\right) \qquad\left( \text{since
}n-\left( n-r\right) =r\right) \\
& =\left( \text{the sum of all principal }r\times r\text{-minors of
}A\right)
\end{align*}
(by the definition of principal minors). This proves Corollary 2.
$\blacksquare$
Lemma 3. Let $A$ be an $n\times n$-matrix over a field $\mathbb{F}$. Let
$\lambda_{1},\lambda_{2},\ldots,\lambda_{n}$ be the eigenvalues of $A$. We
assume that all $n$ of them lie in $\mathbb{F}$. Then, in the polynomial ring
$\mathbb{F}\left[ t\right] $, we have
\begin{equation}
\det\left( tI_n +A\right) =\left( t+\lambda_{1}\right) \left(
t+\lambda_{2}\right) \cdots\left( t+\lambda_{n}\right) .
\end{equation}
Proof of Lemma 3. The eigenvalues of $A$ are defined as the roots of the
characteristic polynomial $\det\left( tI_n -A\right) $ of $A$. (You may be
used to defining the characteristic polynomial of $A$ as $\det\left(
A-tI_n \right) $ instead, but this makes no difference: The polynomials
$\det\left( tI_n -A\right) $ and $\det\left( A-tI_n \right) $ differ
only by a factor of $\left( -1\right) ^{n}$ (in fact, we have $\det\left(
A-tI_n \right) =\left( -1\right) ^{n}\det\left( tI_n -A\right) $), and
thus have the same roots.)
Also, the characteristic polynomial $\det\left( tI_n -A\right) $ of $A$ is
a monic polynomial of degree $n$. And we know that its roots are the
eigenvalues of $A$, which are exactly $\lambda_{1},\lambda_{2},\ldots
,\lambda_{n}$ (with multiplicities). Thus, $\det\left( tI_n -A\right) $ is
a monic polynomial of degree $n$ and has roots $\lambda_{1},\lambda_{2}
,\ldots,\lambda_{n}$. Thus,
\begin{equation}
\det\left( tI_n -A\right) =\left( t-\lambda_{1}\right) \left(
t-\lambda_{2}\right) \cdots\left( t-\lambda_{n}\right)
\end{equation}
(because the only monic polynomial of degree $n$ that has roots $\lambda
_{1},\lambda_{2},\ldots,\lambda_{n}$ is $\left( t-\lambda_{1}\right) \left(
t-\lambda_{2}\right) \cdots\left( t-\lambda_{n}\right) $). Substituting
$-t$ for $t$ in this equality, we obtain
\begin{align*}
\det\left( \left( -t\right) I_n -A\right) & =\left( -t-\lambda
_{1}\right) \left( -t-\lambda_{2}\right) \cdots\left( -t-\lambda
_{n}\right) \\
& =\prod_{i=1}^{n}\underbrace{\left( -t-\lambda_{i}\right) }_{=-\left(
t+\lambda_{i}\right) }=\prod_{i=1}^{n}\left( -\left( t+\lambda_{i}\right)
\right) \\
& =\left( -1\right) ^{n}\underbrace{\prod_{i=1}^{n}\left( t+\lambda
_{i}\right) }_{=\left( t+\lambda_{1}\right) \left( t+\lambda_{2}\right)
\cdots\left( t+\lambda_{n}\right) } \\
& = \left( -1\right) ^{n}\left(
t+\lambda_{1}\right) \left( t+\lambda_{2}\right) \cdots\left(
t+\lambda_{n}\right) .
\end{align*}
Comparing this with
\begin{equation}
\det\left( \underbrace{\left( -t\right) I_n -A}_{=-\left( tI_n
+A\right) }\right) =\det\left( -\left( tI_n +A\right) \right) =\left(
-1\right) ^{n}\det\left( tI_n +A\right) ,
\end{equation}
we obtain
\begin{equation}
\left( -1\right) ^{n}\det\left( tI_n +A\right) =\left( -1\right)
^{n}\left( t+\lambda_{1}\right) \left( t+\lambda_{2}\right) \cdots\left(
t+\lambda_{n}\right) .
\end{equation}
We can divide both sides of this equality by $\left( -1\right) ^{n}$, and
thus obtain $\det\left( tI_n +A\right) =\left( t+\lambda_{1}\right)
\left( t+\lambda_{2}\right) \cdots\left( t+\lambda_{n}\right) $. This
proves Lemma 3. $\blacksquare$
Let us also notice a completely trivial fact:
Lemma 4. Let $\mathbb{F}$ be a field. Let $m$ and $k$ be nonnegative
integers. Let $p\in\mathbb{F}\left[ t\right] $ be a polynomial. Then,
\begin{align*}
& \left( \text{the coefficient of }t^{m+k}\text{ in the polynomial }p\cdot
t^{k}\right) \\
& =\left( \text{the coefficient of }t^{m}\text{ in the polynomial }p\right)
.
\end{align*}
Proof of Lemma 4. The coefficients of the polynomial $p\cdot t^{k}$ are
precisely the coefficients of $p$, shifted to the right by $k$ slots. This
yields Lemma 4. $\blacksquare$
Now we can prove your claim:
Theorem 5. Let $A$ be a diagonalizable $n\times n$-matrix over a field
$\mathbb{F}$. Let $r=\operatorname*{rank}A$. Then,
\begin{align*}
& \left( \text{the product of all nonzero eigenvalues of }A\right) \\
& =\left( \text{the sum of all principal }r\times r\text{-minors of
}A\right) .
\end{align*}
(Here, the product of all nonzero eigenvalues takes the multiplicities of the
eigenvalues into account.)
Proof of Theorem 5. First of all, all $n$ eigenvalues of $A$ belong to
$\mathbb{F}$ (since $A$ is diagonalizable). Moreover, $r=\operatorname*{rank}
A\in\left\{ 0,1,\ldots,n\right\} $ (since $A$ is an $n\times n$-matrix).
The matrix $A$ is diagonalizable; in other words, it is similar to a diagonal
matrix $D\in\mathbb{F}^{n\times n}$. Consider this $D$. Of course, the
diagonal entries of $D$ are the eigenvalues of $A$ (with multiplicities).
Since $A$ is similar to $D$, we have $\operatorname*{rank}
A=\operatorname*{rank}D$. But $D$ is diagonal; thus, its rank
$\operatorname*{rank}D$ equals the number of nonzero diagonal entries of $D$.
In other words, $\operatorname*{rank}D$ equals the number of nonzero
eigenvalues of $A$ (since the diagonal entries of $D$ are the eigenvalues of
$A$). In other words, $r$ equals the number of nonzero eigenvalues of $A$
(since $r=\operatorname*{rank}A=\operatorname*{rank}D$). In other words, the
matrix $A$ has exactly $r$ nonzero eigenvalues.
Label the eigenvalues of $A$ as $\lambda_{1},\lambda_{2},\ldots,\lambda_{n}$
(with multiplicities) in such a way that the first $r$ eigenvalues
$\lambda_{1},\lambda_{2},\ldots,\lambda_{r}$ are nonzero, while the remaining
$n-r$ eigenvalues $\lambda_{r+1},\lambda_{r+2},\ldots,\lambda_{n}$ are zero.
(This is clearly possible, since $A$ has exactly $r$ nonzero eigenvalues.)
Thus, $\lambda_{1},\lambda_{2},\ldots,\lambda_{r}$ are exactly the nonzero
eigenvalues of $A$.
Lemma 3 yields
\begin{align*}
\det\left( tI_n +A\right) & =\left( t+\lambda_{1}\right) \left(
t+\lambda_{2}\right) \cdots\left( t+\lambda_{n}\right) =\prod_{i=1}
^{n}\left( t+\lambda_{i}\right) \\
& =\left( \prod_{i=1}^{r}\left( t+\lambda_{i}\right) \right) \cdot\left(
\prod_{i=r+1}^{n}\left( t+\underbrace{\lambda_{i}}
_{\substack{=0\\\text{(since }\lambda_{r+1},\lambda_{r+2},\ldots,\lambda
_{n}\text{ are zero)}}}\right) \right) \\
& =\left( \prod_{i=1}^{r}\left( t+\lambda_{i}\right) \right)
\cdot\underbrace{\left( \prod_{i=r+1}^{n}t\right) }_{=t^{n-r}}=\left(
\prod_{i=1}^{r}\left( t+\lambda_{i}\right) \right) \cdot t^{n-r}.
\end{align*}
Now, Corollary 2 yields
\begin{align*}
& \left( \text{the sum of all principal }r\times r\text{-minors of
}A\right) \\
& =\left( \text{the coefficient of }t^{n-r}\text{ in the polynomial
}\underbrace{\det\left( tI_n +A\right) }_{=\left( \prod_{i=1}^{r}\left(
t+\lambda_{i}\right) \right) \cdot t^{n-r}}\right) \\
& =\left( \text{the coefficient of }t^{n-r}\text{ in the polynomial }\left(
\prod_{i=1}^{r}\left( t+\lambda_{i}\right) \right) \cdot t^{n-r}\right) \\
& =\left( \text{the coefficient of }t^{0}\text{ in the polynomial }
\prod_{i=1}^{r}\left( t+\lambda_{i}\right) \right) \\
& \qquad\left( \text{by Lemma 4, applied to }m=0\text{ and }k=n-r\text{ and
}p=\prod_{i=1}^{r}\left( t+\lambda_{i}\right) \right) \\
& =\left( \text{the constant term of the polynomial }\prod_{i=1}^{r}\left(
t+\lambda_{i}\right) \right) \\
& =\prod_{i=1}^{r}\lambda_{i}=\lambda_{1}\lambda_{2}\cdots\lambda_{r}\\
& =\left( \text{the product of all nonzero eigenvalues of }A\right)
\end{align*}
(since $\lambda_{1},\lambda_{2},\ldots,\lambda_{r}$ are exactly the nonzero
eigenvalues of $A$). This proves Theorem 5. $\blacksquare$
Note that in the above proof of Theorem 5,
the diagonalizability of $A$ was used only to guarantee that $A$
has exactly $r$ nonzero eigenvalues and that all $n$ eigenvalues of $A$
belong to $\mathbb{F}$.
