TITLE: Beautiful identity: $\sum_{k=m}^n (-1)^{k-m} \binom{k}{m} \binom{n}{k} = \delta_{mn}$
QUESTION [40 upvotes]: Let $m,n\ge 0$ be two integers. Prove that
$$\sum_{k=m}^n (-1)^{k-m} \binom{k}{m} \binom{n}{k} = \delta_{mn}$$
where $\delta_{mn}$ stands for the Kronecker's delta (defined by $\delta_{mn} = \begin{cases} 1, & \text{if } m=n; \\ 0, & \text{if } m\neq n \end{cases}$).
Note: I put the tag "linear algebra" because i think there is an elegant way to attack the problem using a certain type of matrices.
I hope you will enjoy. :)

REPLY [6 votes]: I will try to give an answer using basic complex variables here.
Suppose we are trying to show that
$$\sum_k {n\choose k} (-1)^{k-m} {k\choose m} = \delta_{mn}.$$

Introduce the integral representation
$${k\choose m}
= \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^{k}}{z^{m+1}} \; dz.$$

This gives  for the sum  the integral (the second  binomial coefficent
enforces the range)
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{m+1}} (-1)^m
\sum_{k=0}^n {n\choose k} (-1)^k
(1+z)^k \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{m+1}} (-1)^m
(1-1-z)^n \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{z^n}{z^{m+1}} (-1)^{m+n} \; dz.$$
This integral evaluates to $\delta_{mn}$ by inspection.

We have  not made use of  the properties of complex  integrals here so
this  computation  can  also   be  presented  using  just  algebra  of
generating functions.

Apparently  this method is  due to  Egorychev although  some of  it is
probably folklore.