<s>[INST] How to prove: $\sum_{k=m+1}^{n} (-1)^{k} \binom{n}{k}\binom{k-1}{m}= (-1)^{m+1}$ Show that if $m$ and $n$ are integers with $0\leq m<n$ then $$\sum_{k=m+1}^{n} (-1)^{k} \binom{n}{k}\binom{k-1}{m}= (-1)^{m+1}$$
Attempts:


*

*$(-1)^{k}\binom{n}{k}$ is the coefficient of $x^{k}$ in the expansion of $(1-x)^{n}$

*And $\binom{k-1}{m}$ is the coefficient of $x^{m}$ in the expansion of $(1+x)^{k-1}$. 
Thats all what I could come up with.
 [/INST] For  those who  enjoy integrals  here  is another  approach using  the
Egorychev method as presented in many posts by @FelixMarin and also by
@MarkusScheuer.
Suppose we seek to verify that
$$\sum_{k=m+1}^n (-1)^k {n\choose k} {k-1\choose m} = (-1)^{m+1}.$$
First proof. Introduce
$${k-1\choose m} = {k-1\choose k-1-m} =
\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{k-m}} (1+z)^{k-1} \; dz.$$
Now clearly when  $k\le m$ this vanishes so we may  lower the limit in
the sum to $k=0.$ We obtain
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{z^m}{1+z}
\sum_{k=0}^n {n\choose k} (-1)^k \frac{(1+z)^k}{z^k}  \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{z^m}{1+z}
\left(1-\frac{1+z}{z} \right)^n \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{z^m}{1+z}
\frac{(-1)^n}{z^n} \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(-1)^n}{z^{n-m} (1+z)} \; dz
= (-1)^n (-1)^{n-m-1} = (-1)^{m+1}.$$
Second proof. Introduce
$${k-1\choose m} = {k-1\choose k-1-m} =
\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{m+1}} \frac{1}{(1-z)^{k-m}} \; dz.$$
Now for $1\le k\le m$ this is  $[z^m] (1-z)^{m-k} = 0$ so we may again
extend the summation back to $k=0$,  taking care of the value at $k=0$
which is $(-1)^m.$ We obtain
$$-(-1)^m + \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{m+1}} (1-z)^m 
\sum_{k=0}^n {n\choose k} \frac{(-1)^k}{(1-z)^k}
\; dz
\\ = -(-1)^m + \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{m+1}} (1-z)^m 
\left(1-\frac{1}{1-z}\right)^n
\; dz
\\ = -(-1)^m + \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{m-n+1}} \frac{(-1)^n}{(1-z)^{n-m}} \; dz.$$
Note  however that  we  have $m\lt  n$  so the  contribution from  the
integral vanishes, once more leaving just $$(-1)^{m+1}.$$
</s>