Prove that $\sum_{k=0}^{n} k \binom{n}{k} \binom{m}{k}=n\binom{m+n-1}{n}$
I have tried induction by $n$ and by $m$ with Pascal's Identity, but that always seems to drop $k$ to $k-1$ and I am not sure what to do with that afterwards. I have also tried rewriting the $\binom{m+n-1}{n}$ part to $\sum_{i=o}^{n} \binom{m-2+i}{i}$, but that did not work as well. I know that $\sum_{k=0}^{n} k \binom{n}{k}$ can be obtained from first derivative of $(1+x)^n$ multiplied by $x$ and setting $x=1$ but I don't see how to multiply it then by $\binom{m}{k}$, if this approach is even usable here. I am not really sure what to do with this problem.