combinatorics - Combinatorial identity involving binomial coefficients.

In order to conclude a proof (see last equality in B. Poonen's article), I need to establish the following identity:

$$\forall (\ell,n)\in\mathbb{N}^2,\ell\leqslant n,\sum_{m=\ell}^n{n\choose m}{m\choose \ell}(-1)^{m-\ell}=\left\{\begin{array}{ll}1&\textrm{, if }\ell=n\\0&\textrm{, else}\end{array}\right..$$

I have no clue whether or not this identity is true, although I checked it on the first values of $(\ell,n)$, $(\ell,n)\in\{1,\cdots,15\}^2$.

Any hints will be greatly appreciated!

Answer

First note that

$$\binom{n}m\binom{m}\ell=\binom{n}\ell\binom{n-\ell}{m-\ell}\;,$$

so

$$\begin{align*}

\sum_{m=\ell}^n\binom{n}m\binom{m}\ell(-1)^{m-\ell}&=\sum_{m=\ell}^n\binom{n}\ell\binom{n-\ell}{m-\ell}(-1)^{m-\ell}\\
&=\binom{n}\ell\sum_{m=\ell}^n(-1)^{m-\ell}\binom{n-\ell}{m-\ell}\\
&=\binom{n}\ell\sum_{k=0}^{n-\ell}(-1)^k\binom{n-\ell}k\;.
\end{align*}$$

Now

$$\sum_{k=0}^{n-\ell}(-1)^k\binom{n-\ell}k=\begin{cases}
1,&\text{if }n=\ell\\
0,&\text{otherwise}\;,

\end{cases}$$

and $\dbinom{n}n=1$, so indeed

$$\sum_{m=\ell}^n\binom{n}m\binom{m}\ell(-1)^{m-\ell}=\begin{cases}
1,&\text{if }n=\ell\\
0,&\text{otherwise}\;.
\end{cases}$$