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. :)

The given quantity is the constant term in $ (-1)^m  \times { \sum_{ k \geq 0 } (-1)^{k} \binom{n}{k} (\frac{1}{x})^k } \times \sum_{k \geq 0} \binom{k}{m} x^k $
or the constant term in $(-1)^m \times (1-\frac{1}{x})^n \times x^{m} \times (1-x)^{-(m+1)}$ =  $(-1)^{m+n} x^{m-n} \times (1-x)^{n-m-1} $
If $m >n$ clearly the constant term is 0, if $m < n$ then writing the above as $(-1)^{m+n} \frac{(1-x)^{n-m-1}}{x^{n-m}}$ and noting the maximum exponent of $x$ in numerator is $n-m-1$ we again see the constant term is 0. If $n=m$ then the constant term is clearly 1.
