I tried to verify that $$\sum_{j=m-1}^{n-1}(-1)^{j-m+1}\binom{j}{m-1}\binom{n}{j+1}=1$$ for $n\ge m$.
I believe I can use induction on $n$ and Pascal's formula to justify this; but I wanted to see if there was a way to prove this formula without using induction.
Here is a slightly different variation based upon generating functions. It is convenient to use the coefficient of operator $[z^n]$ to denote the coefficient of $z^n$ of a series. This way we can write e.g. \begin{align*} [z^k](1+z)^n=\binom{n}{k} \end{align*}
We obtain for $1\leq m\leq n$ \begin{align*} \color{blue}{\sum_{j=m-1}^{n-1}}&\color{blue}{(-1)^{j-m+1}\binom{j}{m-1}\binom{n}{j+1}}\\ &=\sum_{j=0}^{n-m}(-1)^{j}\binom{j+m-1}{j}\binom{n}{j+m}\tag{1}\\ &=\sum_{j=0}^{n-m}\binom{-m}{j}\binom{n}{j+m}\tag{2}\\ &=\sum_{j=0}^\infty[z^j](1+z)^{-m}[t^{j+m}](1+t)^n\tag{3}\\ &=[t^m](1+t)^n\sum_{j=0}^\infty t^{-j}[z^j](1+z)^{-m}\tag{4}\\ &=[t^m](1+t)^n\left(1+\frac{1}{t}\right)^{-m}\tag{5}\\ &=[t^0](1+t)^{n-m}\tag{6}\\ &\color{blue}{=1} \end{align*}
Comment:
In (1) we shift the index to start with $j=0$ and apply $\binom{p}{q}=\binom{p}{p-q}$.
In (2) we use the binomial identity $\binom{-p}{q}=\binom{p+q-1}{q}(-1)^q$.
In (3) we apply the coefficient of operator twice and set the upper limit of the series to $\infty$ without changing anything since we are adding zeros only.
In (4) we use the linearity of the coefficient of operator and apply the rule $[z^{p+q}]A(z)=[z^p]z^{-q}A(z)$.
In (5) we apply the substitution rule of the coefficient of operator with $z:=1/t$
\begin{align*}
A(t)=\sum_{j=0}^\infty a_j t^j=\sum_{j=0}^\infty t^j [z^j]A(z)
\end{align*}
In (6) we do some simplifications and are ready to select the coefficient of $[t^0]$.
You can use generating functions: $$(-1)^{j-m+1}\binom{j}{m-1}$$ is the coefficient of $t^j$ in $t^{m-1}/(1+t)^m$ and $$\binom{n}{j+1}$$ is the coefficient of $t^{n-j-1}$ in $(1+t)^n$. Your sum is the coefficient of $t^{n-1}$ in the product of $t^{m-1}/(1+t)^m$ and $(1+t)^n$.
Using Induction
The Cancellation Lemma from this answer says that $$ \sum_{j=m-1}^{n-1}(-1)^{j-m+1}\binom{j+1}{m}\binom{n}{j+1}=[n=m]\tag{1} $$ Then Pascal's Rule implies that $$ \begin{align} a_{n,m} &=\sum_{j=m-1}^{n-1}(-1)^{j-m+1}\binom{j}{m-1}\binom{n}{j+1}\\ &=\sum_{j=m-1}^{n-1}(-1)^{j-m+1}\left[\binom{j+1}{m}-\binom{j}{m}\right]\binom{n}{j+1}\\[9pt] &=[n=m]+a_{n,m+1}\tag{2} \end{align} $$ and the Binomial Theorem says that $$ \begin{align} a_{n,1} &=\sum_{j=0}^{n-1}(-1)^j\binom{j}{0}\binom{n}{j+1}\\ &=\sum_{j=1}^n(-1)^{j-1}\binom{n}{j}\\ &=1-\sum_{j=0}^n(-1)^j\binom{n}{j}\\[6pt] &=1-0^n\\[15pt] &=[n\ge1]\tag{3} \end{align} $$ $(2)$ and $(3)$ combine to give $$ \sum_{j=m-1}^{n-1}(-1)^{j-m+1}\binom{j}{m-1}\binom{n}{j+1}=[n\ge m]\tag{4} $$ where $[\cdots]$ are Iverson Brackets.
Using Generating Functions
$$
\begin{align}
&\sum_{n=0}^\infty\sum_{j=m-1}^{n-1}(-1)^{j-m+1}\binom{j}{m-1}\binom{n}{j+1}x^n\\
&=\sum_{j=m-1}^\infty\sum_{n=j+1}^\infty(-1)^{j-m+1}\binom{j}{m-1}\binom{n}{n-j-1}x^n\tag{5a}\\
&=\sum_{j=m-1}^\infty\sum_{n=j+1}^\infty(-1)^{j-m+1}\binom{j}{m-1}(-1)^{n-j-1}\binom{-j-2}{n-j-1}x^n\tag{5b}\\
&=\sum_{j=m-1}^\infty\sum_{n=0}^\infty(-1)^{n+j-m+1}\binom{j}{m-1}\binom{-j-2}{n}x^{n+j+1}\tag{5c}\\
&=\sum_{j=m-1}^\infty(-1)^{j-m+1}\binom{j}{m-1}\frac{x^{j+1}}{(1-x)^{j+2}}\tag{5d}\\
&=\sum_{j=m-1}^\infty\binom{-m}{j-m+1}\frac{x^{j+1}}{(1-x)^{j+2}}\tag{5e}\\
&=\sum_{j=0}^\infty\binom{-m}{j}\frac{x^{j+m}}{(1-x)^{j+m+1}}\tag{5f}\\
&=\frac{x^m}{(1-x)^{m+1}}\left(\frac1{1-x}\right)^{-m}\tag{5g}\\[6pt]
&=\frac{x^m}{1-x}\tag{5h}
\end{align}
$$
Explanation:
$\text{(5a)}$: switch order of summation and use symmetry of Pascal's Triangle
$\text{(5b)}$: equate with negative binomial coefficient
$\text{(5c)}$: substitute $n\mapsto n+j+1$
$\text{(5d)}$: apply the Binomial Theorem
$\text{(5e)}$: equate with negative binomial coefficient
$\text{(5f)}$: substitute $j\mapsto j+m-1$
$\text{(5g)}$: apply the Binomial Theorem
$\text{(5h)}$: algebra
Equation $(5)$ yields $(4)$ as well.
Here is an alternate proof where the goal was novelty. Start with
$$\sum_{j=m-1}^{n-1} (-1)^{j-m+1} {j\choose m-1} {n\choose j+1} \\ = n \sum_{j=m-1}^{n-1} \frac{(-1)^{j-m+1}}{j+1} {j\choose m-1} {n-1\choose j}.$$
Observe that
$${j\choose m-1} {n-1\choose j} = \frac{(n-1)!}{(n-1-j)! \times (m-1)! (j-m+1)!} \\ = {n-1\choose m-1} {n-m\choose n-1-j}.$$
We get for the sum
$$n {n-1\choose m-1} \sum_{j=m-1}^{n-1} \frac{(-1)^{j-m+1}}{j+1} {n-m\choose n-1-j} \\ = n {n-1\choose m-1} \sum_{j=0}^{n-m} \frac{(-1)^{j}}{j+m} {n-m\choose n-m-j} \\ = n {n-1\choose m-1} \sum_{j=0}^{n-m} \frac{(-1)^{j}}{j+m} {n-m\choose j}.$$
Introduce
$$f(z) = (n-m)! (-1)^{n-m} \frac{1}{z+m} \prod_{q=0}^{n-m} \frac{1}{z-q}.$$
Then we have for $0\le j\le n-m$
$$\mathrm{Res}_{z=j} f(z) = (n-m)! (-1)^{n-m} \frac{1}{j+m} \prod_{q=0}^{j-1} \frac{1}{j-q} \prod_{q=j+1}^{n-m} \frac{1}{j-q} \\ = (n-m)! (-1)^{n-m} \frac{1}{j+m} \frac{1}{j!} \frac{(-1)^{n-m-j}}{(n-m-j)!} = \frac{(-1)^j}{j+m} {n-m\choose j}.$$
Now since $\lim_{R \to\infty} 2\pi R / R^{1+n-m+1} = \lim_{R \to\infty} 2\pi / R^{n-m+1} = 0$ we have $\mathrm{Res}_{z=\infty} f(z) = 0$ and hence
$$\sum_{j=0}^{n-m} \frac{(-1)^{j}}{j+m} {n-m\choose j} = - \mathrm{Res}_{z=-m} f(z) \\ = - (n-m)! (-1)^{n-m} \prod_{q=0}^{n-m} \frac{1}{-m-q} = (n-m)! \prod_{q=0}^{n-m} \frac{1}{m+q} \\ = (n-m)! \frac{(m-1)!}{n!} = \frac{1}{n} {n-1\choose m-1}^{-1}.$$
It follows that our closed form is
$$n {n-1\choose m-1} \times \frac{1}{n} {n-1\choose m-1}^{-1} = 1.$$
