A Binomial Coefficient Sum: $\sum_{m = 0}^{n} (-1)^{n-m} \binom{n}{m} \binom{m-1}{l}$ In my work on $f$-vectors in polytopes, I ran across an interesting sum which has resisted all attempts of algebraic simplification. Does the following binomial coefficient sum simplify?
\begin{align}
\sum_{m = 0}^{n} (-1)^{n-m} \binom{n}{m} \binom{m-1}{l} \qquad l \geq 0
\end{align}
Update: After some numerical work, I believe a binomial sum orthogonality identity is at work here because I see only $\pm 1$ and zeros. Any help would certainly be appreciated.
I take $\binom{-1}{l} = (-1)^{l}$, $\binom{m-1}{l} = 0$ for $0 < m < l$ and the standard definition otherwise. 
Thanks!

$$\sum_{m=0}^n (-1)^{n-m} \binom{n}{m} \binom{m-1}{l} = (-1)^{l+n} + \sum_{l+1 \leq m \leq n} (-1)^{n-m} \binom{n}{m} \binom{m-1}{l}$$
So we need to compute this last sum. It is clearly zero if $l \geq n$, so we assume $l < n$.
It is equal to $f(1)$ where $f(x)= \sum_{l+1 \leq m \leq n} (-1)^{n-m} \binom{n}{m} \binom{m-1}{l} x^{m-1-l}$.
We have that $$\begin{eqnarray*} f(x) & = & \frac{1}{l!} \frac{d^l}{dx^l} \left( \sum_{l+1 \leq m \leq n} (-1)^{n-m} \binom{n}{m} x^{m-1} \right) \\
& = & \frac{1}{l!} \frac{d^l}{dx^l} \left( \frac{(-1)^{n+1}}{x} + \sum_{0 \leq m \leq n} (-1)^{n+1} \binom{n}{m} (-x)^{m-1} \right) \\
& = & \frac{1}{l!} \frac{d^l}{dx^l} \left( \frac{(-1)^{n+1}}{x} + \frac{(x-1)^n}{x} \right) \\
& = & \frac{(-1)^{n+1+l}}{x^{l+1}} + \frac{1}{l!} \sum_{k=0}^l \binom{l}{k} n(n-1) \ldots (n-k+1) (x-1)^{n-k} \frac{(-1)^{l-k} (l-k)!}{x^{1+l-k}}
 \end{eqnarray*}$$
(this last transformation thanks to Leibniz)
and since $n>l$, $f(1)=(-1)^{l+n+1}$.
In the end, your sum is equal to $(-1)^{l+n}$ if $l \geq n$, $0$ otherwise.
