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. :)

You can also use the following finite calculus formula for alternating binomial transforms.  Let $B(f(k),n)$ denote the alternating binomial transform; i.e., $$B(f(k),n) = \sum_{k=0}^n (-1)^k \binom{n}{k} f(k).$$ 
Then (the formula)
$$B(f(k),n) = -B(\Delta f(k),n-1) + f(0)[n=0].$$
(Here, $\Delta f(k)$ is the finite difference $f(k+1) - f(k)$, and the expression $[n=0]$ evaluates to $1$ if $n = 0$ and $0$ otherwise.  Also, note that the first argument to $B$ is a function, while the second is a number.)  
Starting with $f(k) = 1$, we have $\Delta f(k) = 0$.  Since $B(0,n-1)$ is clearly $0$, $B(1,n) = [n=0]$.  The latter is just the known formula 
$$\sum_{k=0}^n (-1)^k \binom{n}{k} = [n=0].$$
Then, continuing to take antidifferences, and using the notation $k^{\underline{m}}$ for the falling factorial $k(k-1)\cdots (k-m+1)$ as well as the power rule for finite differences $\Delta k^{\underline{m}} = m k^{\underline{m-1}}$, we have
$$\sum_{k=0}^n (-1)^k \binom{n}{k}k = -B(1,n-1) = - [n-1=0] = -[n=1],$$
$$\sum_{k=0}^n (-1)^k \binom{n}{k} k^{\underline{2}} = -2B(k,n-1) = 2[n-1 = 1] = 2[n=2],$$
$$\sum_{k=0}^n (-1)^k \binom{n}{k} k^{\underline{3}} = -3B(k^{\underline{2}},n-1) = -6[n-1 = 2] = -6[n=3],$$
and so forth, until we get to 
$$\sum_{k=0}^n (-1)^k \binom{n}{k} k^{\underline{m}} = -mB(k^{\underline{m-1}},n-1) = (-1)^m m![n=m].$$ 
Since $k(k-1)\cdots (k-m+1) = \frac{k!}{(k-m)!}$, dividing both sides of this last identity by $(-1)^m m!$ proves the OP's identity.  
(The finite calculus formula for alternating binomial transforms and this argument are excerpted from my paper "Combinatorial sums and finite differences," Discrete Mathematics 307 (24): 3130-3146, 2007.) 
