Alternating sum over binomial coefficients

Here we have Chu-Vandermonde's Identity in disguise.

Dividing the left hand side by $m!$ we obtain \begin{align*} \color{blue}{\sum_{i=0}^n}&\color{blue}{(-1)^i\binom{k}{n-i}\frac{(m+i)!}{i!m!}}\\ &=\sum_{i=0}^n(-1)^i\binom{k}{n-i}\binom{m+i}{i}\\ &=\sum_{i=0}^n\binom{k}{n-i}\binom{-m-1}{i}\tag{1}\\ &\,\,\color{blue}{=\binom{k-m-1}{n}}\tag{2} \end{align*} and the claim follows.

Comment:

• In (1) we use the binomial identity $\binom{-p}{q}=\binom{p+q-1}{q}(-1)^q$.

• In (2) we apply the Chu-Vandermonde identity.