Looking for combinatorial identity: $\sum\limits_{j=0}^k{n \choose k-j}{m \choose j}$ Is there a nicer closed form expression for the following expression? $$\sum_{j=0}^k{n \choose k-j}{m \choose j}$$

This is Vandermonde's identity:
$$
\sum_{j=0}^k\binom{n}{k-j}\binom{m}{j}=\binom{n+m}{k}
$$
It computes the coefficient of $x^k$ in
$$
(1+x)^n(1+x)^m=(1+x)^{n+m}
$$
