Could I get some help with proving this identity?


$$\sum\_{k=m}^p(-1)^{k+m}\binom{k}{m}\binom{n+p+1}{n+k+1}=\binom{n+p-m}{n}.$$


It has been checked in Matlab for various small $n,m$ and $p$. I have a proof for $m=0$ that involves Pascal's rule to split it into two sums that mostly cancel, but this does not work for all $m$. I have looked at induction on various letters (and looked at simultaneous induction on multiple letters) without any success. 


I would appreciate any help, algebraic or combinatoric.



Binomial ID $\sum_{k=m}^p(-1)^{k+m}\binom{k}{m}\binom{n+p+1}{n+k+1}=\binom{n+p-m}{n}$
Such identities are often reduced to the [Chu--Vandermonde's identity](https://en.wikipedia.org/wiki/Vandermonde%27s_identity) $\sum\_{i+j=\ell} \binom{x}i\binom{y}j=\binom{x+y}\ell$ by using reflection formulae $\binom{x}k=\binom{x}{x-k}$, $\binom{x}k=(-1)^k\binom{k-x-1}k$.


In your case you may write
$$
\sum\_{k=m}^p(-1)^{k+m}\binom{k}{m}\binom{n+p+1}{n+k+1}=
\sum\_{k=m}^p \binom{-m-1}{k-m}\binom{n+p+1}{p-k}=
\binom{n+p-m}{p-m}
$$
as you need, so it is Chu--Vandermonde for $x=-m-1$, $y=n+p+1$, $\ell=p-m$.


