Chu-Vandermonde Identity/Proof 4

Proof
From Sum of Products of $\displaystyle {r - t k \choose k} {s + t \left({n - k}\right) \choose n - k} \frac r {r - t k}$:

where $r, s, t \in \R, n \in \Z$.

Setting $t = 0$:
 * $\displaystyle \sum_{k \mathop \ge 0} \binom r k \binom s {n - k} = \binom {r + s} n$

which is the result required.