I need to show that $\sum_{k=m+1}^{n} (-1)^k \binom {n} {k} \binom {k-1} {m} = (-1)^{m+1}$.
What I've done so far is divide by $(-1)^{m+1}$, and so our expression takes the form of: $\sum_{k=m+1}^{n} (-1)^{k-m-1} \binom {n} {k} \binom {k-1} {m}$. I thought of substituting $k-m-1$ with a variable, let's say $p$. Then, we'd have $\sum_{p=0}^{n-(m+1)} (-1)^{p} \binom {n} {k} \binom {k-1} {p} = 1$. (since $\binom {k-1} {m} = \binom {k-1}{k-1-m}$ where $k-m-1=p$
I'm stuck here though. I tried to find a way to see if I could simplify the multiplication/binomials, but it doesn't seem to lead to anywhere good.
Any suggestions are welcomed. Thank you!
Here's a way to do it (I link to an identity at the end, which you may know, or may need to prove):
$$\sum_{k=m+1}^{n} (-1)^k \binom {n} {k} \binom {k-1} {m} = \sum_{k=m+1}^{n} (-1)^k \dfrac{n!}{(n-k)!k!}\dfrac{(k-1)!}{(k-1-m)!m!} $$
Since $k>m$ let $p=k-m$. Then
$$ \dfrac{n!}{(n-k)!k!}\dfrac{(k-1)!}{(k-1-m)!m!}= \dfrac{p}{k}\dfrac{n!}{m!(n-m)!}\dfrac{(n-m)!}{(n-m-p)!p!}=\dfrac{p}{p+m} {n\choose m} {n-m\choose p}$$
So
$$\sum_{k=m+1}^{n} (-1)^k \binom {n} {k} \binom {k-1} {m} = (-1)^{m+1}{n\choose m}\sum_{p=1}^{(n-m)}\dfrac{p}{p+m} (-1)^{p-1} {n-m\choose p} $$
By Partial Fraction Decomposition of $\dfrac{1}{{n\choose m}}$, we have
$$\dfrac{1}{{n\choose m}} = \dfrac{1}{{n\choose n-m}} = \sum_{p=1}^{(n-m)}\dfrac{p}{p+m} (-1)^{p-1} {n-m\choose p} $$
Proving the proposition.