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!

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$\ds{\sum_{m = 0}^{n}\pars{-1}^{n - m}{n \choose m}{m - 1 \choose \ell}:\
     {\large ?}.\qquad\ell \geq 0}$

\begin{align}
&\color{#66f}{\large\sum_{m = 0}^{n}\pars{-1}^{n - m}{n \choose m}
{m - 1 \choose \ell}}
\\[3mm]&=\pars{-1}^{n}\sum_{m = 0}^{n}\pars{-1}^{m}{n \choose m}
\oint_{0\ <\ \verts{z}\ =\ a\ <\ 1}{\pars{1 + z}^{m - 1} \over z^{\ell + 1}}
\,{\dd z \over 2\pi\ic}
\\[3mm]&=\pars{-1}^{n}\oint_{0\ <\ \verts{z}\ =\ a\ <\ 1}
{1 \over z^{\ell + 1}\pars{1 + z}}
\sum_{m = 0}^{n}{n \choose m}\pars{-z - 1}^{m}\,{\dd z \over 2\pi\ic}
\\[3mm]&=\pars{-1}^{n}\oint_{0\ <\ \verts{z}\ =\ a\ <\ 1}
{1 \over z^{\ell + 1}\pars{1 + z}}
\bracks{1 + \pars{-z - 1}}^{n}\,{\dd z \over 2\pi\ic}
\\[3mm]&=\oint_{0\ <\ \verts{z}\ =\ a\ <\ 1}{1 \over z^{\ell - n + 1}\pars{1 + z}}
{\dd z \over 2\pi\ic}
=\sum_{k = 0}^{\infty}\pars{-1}^{k}\oint_{0\ <\ \verts{z}\ =\ a\ <\ 1}{1 \over z^{\ell - n - k + 1}}{\dd z \over 2\pi\ic}
\\[3mm]&=\sum_{k = 0}^{\infty}\pars{-1}^{k}\,\delta_{\ell - n,k}
=\color{#66f}{\large\left\lbrace\begin{array}{lcl}
\pars{-1}^{\ell - n} & \mbox{if} & \ell \geq n
\\[2mm]
0&&\mbox{otherwise}
\end{array}\right.}
\end{align}

