Show that $ \sum_k \binom{r}{k} \binom{s+k}{n} (-1)^k = (-1)^r \binom{s}{n-r} $ I can't resolve this exercise and I need tips.
Let be $n$ integer, $s$ real and $r \geq 0$ integer. Show that
$$ \sum_k \binom{r}{k} \binom{s+k}{n} (-1)^k = (-1)^r \binom{s}{n-r} $$
 A: Vandermonde's Identity says
$$
\binom{s+k}{n}=\sum_{j=0}^n\binom{s}{n-j}\binom{k}{j}\tag{1}
$$
Thus,
$$
\begin{align}
\sum_{k=0}^r(-1)^k\binom{r}{k}\binom{s+k}{n}
&=\sum_{k=0}^r(-1)^k\binom{r}{k}\sum_{j=0}^n\binom{s}{n-j}\binom{k}{j}\tag{2}\\
&=\sum_{k=0}^r\sum_{j=0}^n(-1)^k\binom{r}{k}\binom{k}{j}\binom{s}{n-j}\tag{3}\\
&=\sum_{k=0}^r\sum_{j=0}^n(-1)^k\binom{r}{j}\binom{r-j}{k-j}\binom{s}{n-j}\tag{4}\\
&=(-1)^r\binom{s}{n-r}\tag{5}
\end{align}
$$
Explanation:
$(2)$: apply $(1)$
$(3)$: rearrange terms
$(4)$: $\binom{r}{k\vphantom{j}}\binom{k}{j}=\binom{r}{j}\binom{r-j}{k-j}$ (just write out the coefficients in terms of factorials)
$(5)$: $\sum\limits_{k=0}^r(-1)^k\binom{r-j}{k-j}=\sum\limits_{k=0}^{r-j}(-1)^{k+j}\binom{r-j}{k}=(-1)^j(1-1)^{r-j}=(-1)^j[j=r]$
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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$\ds{\sum_{k}{r \choose k}{s + k \choose n}\pars{-1}^{k}
    = \pars{-1}^{r}{s \choose n - r}:\ {\large ?}}.\quad$
$\ds{n\ \mbox{integer}, s\ \mbox{real and}\  r\ \mbox{an integer}\ \geq 0}$.

\begin{align}
&\color{#66f}{\large\sum_{k}{r \choose k}{s + k \choose n}\pars{-1}^{k}}
=\sum_{k}{r \choose k}\pars{-1}^{k}\oint_{\verts{z}\ =\ 1}{\pars{1 + z}^{s + k} \over z^{n + 1}}
\,{\dd z \over 2\pi\ic}
\\[3mm]&=\oint_{\verts{z}\ =\ 1}{\pars{1 + z}^{s} \over z^{n + 1}}
\sum_{k}{r \choose k}\pars{-1 - z}^{k}\,{\dd z \over 2\pi\ic}
=\oint_{\verts{z}\ =\ 1}{\pars{1 + z}^{s} \over z^{n + 1}}
\bracks{1 + \pars{-1 - z}}^{r}\,{\dd z \over 2\pi\ic}
\\[3mm]&=\pars{-1}^{r}\oint_{\verts{z}\ =\ 1}{\pars{1 + z}^{s} \over z^{n - r + 1}}
\,{\dd z \over 2\pi\ic}
=\color{#66f}{\large\pars{-1}^{r}{s \choose n - r}}
\end{align}

