$\sum_{k=m}^{n}(-1)^k\binom{n}{k}\binom{k}{m}=0, n>m\geq 0$ I got quite some trouble trying to prove this. 
$$\sum_{k=m}^{n}(-1)^k\binom{n}{k}\binom{k}{m}=0, n>m\geq 0$$
I tried using $$\binom{n}{m}\binom{m}{k}=\binom{n}{k}\binom{n-k}{m-k}$$ and then changing the summation boundaries as $k=0, n=n-m$, but that took me nowhere... I would really appreciate help. Thank you. 
 A: Here is the proof.
\begin{align}
\sum_{k=m}^{n}(-1)^k\binom{n}{k}\binom{k}{m} &= \sum_{k=m}^{n}(-1)^k
\frac{n!}{k!(n-k)!}\frac{k!}{(k-m)!m!} \\
&= \sum_{k=m}^{n}(-1)^k
\frac{n!}{(n-m)!m!}\frac{(n-m)!}{(n-k)!(k-m)!} \\
&= \sum_{k=m}^{n}(-1)^k \binom{n}{m}\binom{n-m}{k-m} \\
&= \binom{n}{m} \sum_{k=m}^{n}(-1)^k \binom{n-m}{k-m} \\
&= (-1)^m \binom{n}{m} \sum_{j=0}^{n-m} \binom{n-m}{j}(-1)^{j}(1)^{n-m-j} \\
&= (-1)^m \binom{n}{m} (1-1)^{n-m} = 0.
\end{align}
Note that in the fifth equality, I set $j = k -m$ and used $(-1)^{j+m} = (-1)^j(-1)^m$.  
A: We seek for $n\gt m\ge 0$ that
$$\sum_{k=m}^n {n\choose k} (-1)^k {k\choose m} = 0.$$
Using coefficient extractors the sum becomes
$$\sum_{k=m}^n {n\choose k} (-1)^k
[z^m] (1+z)^k
\\ = [z^m] \sum_{k=m}^n {n\choose k} (-1)^k
(1+z)^k.$$
We may lower $k$ to zero because  when $k\lt m$ we have $[z^m] (1+z)^k
= 0.$ We find
$$[z^m] \sum_{k=0}^n {n\choose k} (-1)^k (1+z)^k
\\ = [z^m] (1-(1+z))^n
= [z^m] (-1)^n z^n = (-1)^n \delta_{m,n}.$$
This is the claim.
