Sum over k of r Choose k by s+k Choose n by -1^r-k/Proof 1

Proof
The proof proceeds by induction on $r$.

For all $r \in \Z_{>0}$, let $P \left({r}\right)$ be the proposition:
 * $\displaystyle \sum_k \binom r k \binom {s + k} n \left({-1}\right)^{r - k} = \binom s {n - r}$

Basis for the Induction
$P \left({0}\right)$ is the case:

Thus $P \left({0}\right)$ is seen to hold.

This is the basis for the induction.

Induction Hypothesis
Now it needs to be shown that, if $P \left({m}\right)$ is true, where $m \ge 0$, then it logically follows that $P \left({m + 1}\right)$ is true.

This is the induction hypothesis:
 * $\displaystyle \sum_k \binom m k \binom {s + k} n \left({-1}\right)^{m - k} = \binom s {n - m}$

from which it is to be shown that:
 * $\displaystyle \sum_k \binom {m + 1} k \binom {s + k} n \left({-1}\right)^{m + 1 - k} = \binom s {n - \left({m + 1}\right)}$

Induction Step
This is the induction step:

So $P \left({m}\right) \implies P \left({m + 1}\right)$ and the result follows by the Principle of Mathematical Induction.

Therefore:
 * $\displaystyle \sum_k \binom r k \binom {s + k} n \left({-1}\right)^{r - k} = \binom s {n - r}$

for all $s \in \R, r \in \Z_{\ge 0}, n \in \Z$.