Looking for a proof of two combinatorial summation identities

Using a coefficient extractor Eg \begin{eqnarray*} \binom{k}{m} = [x^m]:(1+x)^k. \end{eqnarray*} We have \begin{eqnarray*} \sum_{k\ge1}\binom km\binom{n-1}{k-1} &=& [x^m]: \sum_{k=1}^{n} \binom{n-1}{k-1} (1+x)^k \\ &=& [x^m]: (1+x) (2+x)^{n-1} \\ &=& 2^{n-1-m} \binom{n-1}{m} +2^{n-m} \binom{n-1}{m-1} \\ &=& 2^{n-1-m}\left( \binom{n-1}{m} +2 \binom{n-1}{m-1} \right)\\ &=& 2^{n-1-m} \frac{(n-1)!}{m!(n-m)!} \left( n-m+2m \right)\\ &=& 2^{n-1-m} \frac{(n-1)!}{m!(n-m)!} (n+m).\\ \end{eqnarray*} Which is the first identity (upto a factor of $n$).