The following is a detailed proof following the hint posted as an answer by
@user26857. We will use the
following fact:
Lemma 1. Let $K$ be a field. Let $L$ be a $K$-algebra that is
finite-dimensional as a $K$-vector space. Let $R$ be a $K$-subalgebra of $L$
that is an integral domain. Then, $R$ is a field.
Lemma 1 is the particular case of the Theorem 1 from
https://math.stackexchange.com/a/3161397/ when $g$ is an inclusion map. (Of
course, this particular case is easily seen to be equivalent to the general case.)
Let us now solve your question.
Assume that $y_{1},y_{2},\ldots,y_{r}$ are $r$ elements of $E$ that are
linearly independent over $F$. We must show that the $r$ elements $y_{1}
^{p},y_{2}^{p},\ldots,y_{r}^{p}$ of $E$ are linearly independent over $F$ as well.
The list $\left( y_{1},y_{2},\ldots,y_{r}\right) $ of vectors in $E$ is
linearly independent over $F$; thus, we can extend this list to a basis
$\left( y_{1},y_{2},\ldots,y_{n}\right) $ of the $F$-vector space $E$ (since
$E$ is a finite-dimensional $F$-vector space). Consider such a basis $\left(
y_{1},y_{2},\ldots,y_{n}\right) $. Since $\left( y_{1},y_{2},\ldots
,y_{n}\right) $ is a basis of $E$, there is a unique endomorphism $\phi$ of
the $F$-vector space $E$ that sends $y_{1},y_{2},\ldots,y_{n}$ to $y_{1}
^{p},y_{2}^{p},\ldots,y_{n}^{p}$, respectively. Consider this $\phi$.
The image $\phi\left( E\right) $ of $E$ is an $F$-vector subspace of $E$
(since $\phi$ is $F$-linear), but it has further properties. To wit, let us
first show that $E^{p}\subseteq\phi\left( E\right) $.
Indeed, let $w\in E^{p}$. Then, $w=e^{p}$ for some $e\in E$. Consider this
$e$. Then, $e=\sum_{i=1}^{n}a_{i}y_{i}$ for some $a_{1},a_{2},\ldots,a_{n}\in
F$ (since $\left( y_{1},y_{2},\ldots,y_{n}\right) $ is a basis of the
$F$-vector space $E$). Consider these $a_{1},a_{2},\ldots,a_{n}$. Now,
\begin{align*}
w & =e^{p}=\left( \sum_{i=1}^{n}a_{i}y_{i}\right) ^{p}\qquad\left(
\text{since }e=\sum_{i=1}^{n}a_{i}y_{i}\right) \\
& =\sum_{i=1}^{n}a_{i}^{p}y_{i}^{p}
\end{align*}
(since we are in characteristic $p>0$, so that the map $E\rightarrow
E,\ z\mapsto z^{p}$ is a ring endomorphism of $E$). On the other hand, the
definition of $\phi$ yields
\begin{align*}
\phi\left( \sum_{i=1}^{n}a_{i}^{p}y_{i}\right) =\sum_{i=1}^{n}a_{i}^{p}
y_{i}^{p}
\end{align*}
(since $a_{1},a_{2},\ldots,a_{n}$ are scalars in $F$). Comparing these two
equalities, we find $w=\phi\left( \sum_{i=1}^{n}a_{i}^{p}y_{i}\right)
\in\phi\left( E\right) $.
Forget that we fixed $w$. We thus have shown that $w\in\phi\left( E\right) $
for each $w\in E^{p}$. In other words, $E^{p}\subseteq\phi\left( E\right) $.
Hence, $1\in E^{p}\subseteq\phi\left( E\right) $, so that $F=F\cdot
1\subseteq\phi\left( E\right) $ (since $\phi\left( E\right) $ is an
$F$-vector subspace of $E$ and since $1\in\phi\left( E\right) $).
Next, we shall show that the set $\phi\left( E\right) $ is closed under
multiplication. Indeed, $\phi\left( E\right) $ is the image of the
$F$-linear map $\phi$, which sends the basis elements $y_{1},y_{2}
,\ldots,y_{n}$ to $y_{1}^{p},y_{2}^{p},\ldots,y_{n}^{p}$, respectively. Thus,
$\phi\left( E\right) $ is the $F$-linear span of the elements $y_{1}
^{p},y_{2}^{p},\ldots,y_{n}^{p}$. Hence, in order to show that $\phi\left(
E\right) $ is closed under multiplication, it will suffice to prove that the
pairwise products $y_{i}^{p}y_{j}^{p}$ of the latter elements $y_{1}^{p}
,y_{2}^{p},\ldots,y_{n}^{p}$ belong to $\phi\left( E\right) $. But this is
easy: For any $i,j\in\left\{ 1,2,\ldots,n\right\} $, we have $y_{i}^{p}
y_{j}^{p}=\left( y_{i}y_{j}\right) ^{p}\in E^{p}\subseteq\phi\left(
E\right) $. Thus, we have shown that $\phi\left( E\right) $ is closed under
multiplication. Since $\phi\left( E\right) $ is furthermore an $F$-vector
subspace of $E$ and contains $1$ (since $1\in\phi\left( E\right) $), we thus
conclude that $\phi\left( E\right) $ is an $F$-subalgebra of $E$. In
particular, $\phi\left( E\right) $ is a subring of the field $E$, and thus
is an integral domain (since any subring of a field is an integral domain).
Hence, Lemma 1 (applied to $K=F$ and $L=E$ and $R=\phi\left( E\right) $)
yields that $\phi\left( E\right) $ is a field. Therefore, $\phi\left(
E\right) $ is a subfield of $E$ that contains both $F$ and $E^{p}$ as
subfields (since $F\subseteq\phi\left( E\right) $ and $E^{p}\subseteq
\phi\left( E\right) $).
However, $F\left( E^{p}\right) $ is the smallest such subfield (by the
definition of $F\left( E^{p}\right) $). Hence, $F\left( E^{p}\right)
\subseteq\phi\left( E\right) $. Now, recall that $E=F\left( E^{p}\right)
$. Hence, $E=F\left( E^{p}\right) \subseteq\phi\left( E\right) $.
Combining this with $\phi\left( E\right) \subseteq E$, we obtain
$E=\phi\left( E\right) $. In other words, the map $\phi$ is surjective.
However, a surjective endomorphism of a finite-dimensional $F$-vector space
must necessarily be injective. Thus, $\phi$ is injective (since $\phi$ is a
surjective endomorphism of the finite-dimensional $F$-vector space $E$). Since
$\phi$ is the $F$-linear map that sends the basis vectors $y_{1},y_{2}
,\ldots,y_{n}$ to $y_{1}^{p},y_{2}^{p},\ldots,y_{n}^{p}$, this entails that
the vectors $y_{1}^{p},y_{2}^{p},\ldots,y_{n}^{p}$ are $F$-linearly
independent. Hence, the vectors $y_{1}^{p},y_{2}^{p},\ldots,y_{r}^{p}$ are
$F$-linearly independent (since $n\geq r$). Qed.
