Proof that $\sum\limits_k(-1)^k{r \choose m + k}{s+k \choose n}=(-1)^{r+n}{s-m \choose n-r}$ $\sum\limits_k(-1)^k{r \choose m + k}{s+k \choose n}=(-1)^{r+n}{s-m \choose n-r}$
$\sum\limits_k$ - means for all possible $k$
I think we should try induction there
 A: $$\begin{align}
&\;\;\;\sum_k(-1)^k \binom{r}{m+k}\binom{s+k}n\\
&=\sum_k (-1)^k \binom{r}{m+k}\binom{s+k}{s+k-n}\\
&=\sum_k (-1)^k \binom{r}{m+k}(-1)^{s+k-n}\binom{-n-1}{s+k-n}\\
&=(-1)^{s-n} \sum_k \binom{r}{r-m-k}\binom{-n-1}{s-n+k}\\
&=(-1)^{s-n}\binom{r-n-1}{r-m+s-n}\\
&=(-1)^{s-n}(-1)^{r-m+s-n}\binom{-m+s}{r-m+s-n}\\
&=(-1)^{r-m}\binom{s-m}{s-m+r-n}\\
&=(-1)^{r+m}\binom{s-m}{n-r}
\end{align}$$
The above uses:
(1) Vandermonde Identity
(2) Upper Negation
A: In evaluating
$$\sum_k (-1)^k {r\choose m+k} {s+k\choose n}$$
we notice that ${r\choose m+k}$ is zero when $k\lt -m.$ We must also have
$r\ge m+k$ or $k\le r-m$ because $r^\underline{m+k} = 0$ otherwise.
We get
$$\sum_{k=-m}^{r-m} (-1)^k {r\choose m+k} {s+k\choose n}
\\ = (-1)^m \sum_{k=0}^{r}
(-1)^k {r\choose k} {s-m+k\choose n}
\\ = (-1)^m \sum_{k=0}^{r}
(-1)^k {r\choose k} [z^n] (1+z)^{s-m+k}
\\ = [z^n] (1+z)^{s-m} (-1)^m \sum_{k=0}^{r}
(-1)^k {r\choose k}  (1+z)^{k}
\\ = [z^n] (1+z)^{s-m} (-1)^m
(1-(1+z))^r
\\ = [z^n] (1+z)^{s-m} (-1)^{m+r} z^r
= (-1)^{m+r} [z^{n-r}] (1+z)^{s-m} 
\\ = (-1)^{m+r} {s-m\choose n-r}.$$
