Identity involving double sum with factorials In the course of a calculation, I have met a complicated identity, which I want to prove.
Let $m>0$ and $0<\ell<m$ be integers. Let $(x)^{(n)}=x(x+1)\cdots (x+n-1)$ be the rising factorial. Then
$$\sum_{n=1}^m\sum_{k=0}^{n-1}{m \choose n}(-1)^{n-k}\frac{(n+m-k-\ell-2)!(k+\ell)!}{(n-k-1)!^2k!^2}(x-k)^{(n)}=(-1)^{m-\ell}(x-\ell)^{(m)}.$$
I have tried writing everything in terms of powers of $x$, but didn't really help me.
How to prove this identity?
Version with binomials:
$$\sum_{n=1}^m\sum_{k=0}^{n-1}(-1)^{n-k}{m \choose n}{n+m-k-\ell-2 \choose n-k-1}{k+\ell\choose k}{n-1\choose k}\frac{(x-k)^{(n)}}{(n-1)!}=(-1)^{m-\ell}{m-1\choose \ell}\frac{(x-\ell)^{(m)}}{(m-1)!}.$$
Or, defining
$$F_{n,k}(x)=(-1)^{n-k}{n-1\choose k}\frac{(x-k)^{(n)}}{(n-1)!},$$
we can write
$$\sum_{n=1}^m\sum_{k=0}^{n-1}{m \choose n}{n+m-k-\ell-2 \choose n-k-1}{k+\ell\choose k}F_{n,k}(x)=F_{m,\ell}(x).$$
 A: Re-writing slightly we find
$$\sum_{n=1}^m {m\choose n} n
\sum_{k=0}^{n-1} (-1)^{n-k}
{n+m-k-\ell-2\choose n-k-1}
{k+\ell\choose \ell}
{n-1\choose k} {x-k+n-1\choose n}
= m (-1)^{m-\ell} {m-1\choose \ell}
{x-\ell + m-1\choose m}.$$
We may treat this as polynomials in $x$ and take it to be a positive
integer. It then generalizes to complex $x.$ Working with the inner sum
we find
$$[z^{n-1}] (1+z)^{n+m-\ell-2}
[w^\ell] (1+w)^\ell
[v^n] (1+v)^{x+n-1}
\\ \times
\sum_{k=0}^{n-1} (-1)^{n-k} {n-1\choose k} \frac{z^k}{(1+z)^k}
(1+w)^k (1+v)^{-k}
\\ = (-1)^n [z^{n-1}] (1+z)^{n+m-\ell-2}
[w^\ell] (1+w)^\ell
[v^n] (1+v)^{x+n-1}
\\ \times
\left[ 1-\frac{z(1+w)}{(1+z)(1+v)}\right]^{n-1}
\\ = (-1)^n [z^{n-1}] (1+z)^{m-\ell-1}
[w^\ell] (1+w)^\ell
[v^n] (1+v)^{x}
(1+v+vz-wz)^{n-1}.$$
Using $q$ as the index variable we get for the outer sum
$$m \sum_{q=1}^m {m-1\choose q-1}
(-1)^q [z^{q-1}] (1+z)^{m-\ell-1}
\\ \times [w^\ell] (1+w)^\ell
[v^q] (1+v)^{x}
(1+v+vz-wz)^{q-1}
\\ = m \sum_{q=0}^{m-1} {m-1\choose q}
(-1)^{m-q} [z^{m-1}] z^q (1+z)^{m-\ell-1}
\\ \times [w^\ell] (1+w)^\ell
[v^m] v^q (1+v)^{x}
(1+v+vz-wz)^{m-q-1}
\\ = m [z^{m-1}] (1+z)^{m-\ell-1}
[v^m] (1+v)^{x} [w^\ell] (1+w)^\ell
\\ \times \sum_{q=0}^{m-1} {m-1\choose q}
(-1)^{m-q} z^q
v^q (1+v+vz-wz)^{m-q-1}
\\ = - m [z^{m-1}] (1+z)^{m-\ell-1}
[v^m] (1+v)^{x} [w^\ell] (1+w)^\ell
(wz-1-v)^{m-1}.$$
Expanding the last powered term we obtain
$$- m [z^{m-1}] (1+z)^{m-\ell-1}
[v^m] (1+v)^{x} [w^\ell] (1+w)^\ell
\\ \times \sum_{p=0}^{m-1} {m-1\choose p} w^p z^p
(-1)^{m-1-p} (1+v)^{m-1-p}.$$
For the coefficient extractor in $z$ to return a non-zero value we must
have $m-1-p \le m-\ell-1$ (note that with $0\lt \ell\lt m$ the term
$(1+z)^{m-\ell-1}$ is finite). This says that $\ell \le p.$ On the
other hand the coefficient extractor in $w$ requires $p\le \ell$
(the term $(1+w)^\ell$ is finite as well and we use the residue
definition ${\ell\choose \ell-p} = \; \underset{w}{\mathrm{res}}\;
\frac{1}{w^{\ell-p+1}} (1+w)^\ell.$) The only $p$ to fulfill both
conditions is $p=\ell$ and we find
$$- m [z^{m-1}] (1+z)^{m-\ell-1}
[v^m] (1+v)^{x} [w^\ell] (1+w)^\ell
\\ \times  {m-1\choose \ell} w^\ell z^\ell
(-1)^{m-1-\ell} (1+v)^{m-1-\ell}
\\ = -m {m-\ell-1\choose m-\ell-1} {x+m-1-\ell\choose m}
{\ell\choose \ell-\ell}
(-1)^{m-1-\ell} {m-1\choose \ell}.$$
This at last simplifies to
$$m (-1)^{m-\ell} {m-1\choose \ell}
{x-\ell+m-1\choose m}$$
which is the claim.
