Beautiful identity: $\sum_{k=m}^n (-1)^{k-m} \binom{k}{m} \binom{n}{k} = \delta_{mn}$ 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. :)

This follows easily from the Multinomial Theorem, I believe.
$$ 1 = 1^n = (1 - x + x)^n$$
$$ = \sum_{a+b+c=n} {n \choose a,b,c} 1^a \cdot (-x)^b \cdot x^c$$
 $$ = \sum_{m=0}^{n} \sum_{k=m}^{n} {n \choose m,k-m,n-k} 1^{m}  \cdot (-x)^{k-m} \cdot x^{n-k} $$
  $$ = \sum_{m=0}^{n} \left[ \sum_{k=m}^{n} (-1)^{k-m} {k \choose m}{n \choose k} \right] x^{n-m}$$
Comparing coefficients now gives the result immediately.
