A binomial identity I was wandering if someone knows an elementary proof of the following identity:
$$
\frac{(a)_n (b)_n}{(n!)^2} = \sum_{k=0}^n (-1)^k {1-a-b \choose k} 
\frac{(1-a)_{n-k}(1-b)_{n-k}}{((n-k)!)^2}\ ,
$$
where $a,b$ are arbitrary real numbers,
$$(x)_0=1,\quad 
(x)_n:=x(x+1)\cdots(x+n-1) \quad 
\mbox{for $n\geq 1$}
$$ 
is the Pochhammer's symbol, and 
$$
{x\choose 0}=1,\quad 
{x\choose k} = \frac{x(x-1)\cdots (x-k+1)}{k!} \quad 
\mbox{for $k\geq 1$}$$
is a binomial coefficient.
The proof that I know uses the Hypergeometric differential equation. One has to continue analytically the solutions along a path connecting two singular points. This could be done by some well known integral representation of the Hypergeometric function. 
I think that there should be a combinatorial proof. Since this is an identity between polynomials in $a$ and $b$, it is enough to prove it for $a$ and $b$ negative integers, i.e., we may assume that $a=-p$ and $b=-q$ where $p,q\in \mathbb{Z}_{\geq 0}$. The identity then turns into
$$
{p\choose n}{q\choose n} = 
\sum_{k=0}^n (-1)^k {1+p+q\choose k}{p+n-k\choose n-k}{q+n-k\choose n-k}.
$$
I did not try very hard to proof the above identity and I did not search the literature that much, but since it comes from an interesting subject I think that it is worth finding an alternative proof. 
 A: Here's a proof using generating functions.
We'll use the following repeatedly: $\displaystyle \sum_{k \ge 0} \binom{n}{k} x^k = (1+x)^n, \sum_{k \ge 0} \binom{k}{n} x^k = \frac{x^n}{(1-x)^{n+1}}$
We want to prove the following: 
$$
{p\choose n}{q\choose n} = 
\sum_{k=0}^n (-1)^k {1+p+q\choose k}{p+n-k\choose n-k}{q+n-k\choose n-k}.
$$
Let's attack the gullible left hand side first. Multiply by $x^n y^q$ and sum over all $n, q \ge 0$, and rearrange the summations to get
$$\sum_{n \ge 0} {p\choose n} x^n \sum_{q \ge 0} {q\choose n} y^q = \sum_{n \ge 0} {p\choose n} x^n \frac{y^n}{(1-y)^{n+1}} = \frac{1}{1-y} \left(1 + \frac{xy}{1-y} \right)^p = \frac{(1-y +xy)^p}{(1-y)^{p+1}}$$
So much for the left hand side. Now to the right hand side.
Firstly, change $k \rightarrow n-k$ in the summation, so that our right hand side becomes: 
$$\sum_{k=0}^n (-1)^{n-k} {1+p+q\choose n-k}{p+k\choose k}{q+k\choose k} = \sum_{k=0}^n (-1)^{n-k} {1+p+q\choose n-k}{p+k\choose p}{q+k\choose k}$$
Now, multiply first by $x^n$, sum over $n \ge 0$, and rearrange to get : (We'll multiply $y^q$ later)
$$\sum_{k \ge 0} {p+k\choose p}{q+k\choose k} \sum_{n \ge 0} (-1)^{n-k} {1+p+q\choose n-k} x^n = \sum_{k \ge 0} {p+k\choose p}{q+k\choose k} x^k (1-x)^{1+p+q}$$
Now comes $y$. Multiply by $y^q$, sum over $q \ge 0$, and rearrange to get :
$$(1-x)^{1+p} \sum_{k \ge 0} {p+k\choose p} x^k \sum_{q \ge 0} {q+k\choose k} y^q (1-x)^{q} = (1-x)^{1+p} \sum_{k \ge 0} {p+k\choose p} \frac{x^k}{(1-y+xy)^{k+1}}$$
Now, this is equal to $\displaystyle \frac{(1-x)^{1+p}}{(1-y+xy)} \frac{1}{(1-t)^{1+p}}$, where $\displaystyle t = \frac{x}{(1-y+xy)} \implies 1-t = \frac{(1-x)(1-y)}{(1-y+xy)}$, giving us that $\displaystyle \frac{(1-x)^{1+p}}{(1-y+xy)} \frac{1}{(1-t)^{1+p}} = \frac{(1-y +xy)^p}{(1-y)^{p+1}}$, which was exactly what was needed.
A: Suppose we seek to evaluate
$$\sum_{k=0}^n (-1)^k
{p+q+1\choose k} {p+n-k\choose n-k} {q+n-k\choose n-k}$$ which is
claimed to be $${p\choose n}{q\choose n}.$$
Introduce $${p+n-k\choose n-k}
= \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{p+n-k}}{z^{n-k+1}} 
\; dz$$
and $${q+n-k\choose n-k}
= \frac{1}{2\pi i}
\int_{|w|=\gamma} \frac{(1+w)^{q+n-k}}{w^{n-k+1}} 
\; dw.$$
Observe that these integrals vanish when $k\gt n$ and we may extend
$k$ to infinity.
We thus obtain for the sum $$\frac{1}{2\pi i}
\int_{|z|=\epsilon} 
\frac{(1+z)^{p+n}}{z^{n+1}} 
\frac{1}{2\pi i}
\int_{|w|=\gamma} 
\frac{(1+w)^{q+n}}{w^{n+1}} 
\\ \times \sum_{k\ge 0} {p+q+1\choose k} (-1)^k
\frac{z^k w^k}{(1+z)^k (1+w)^k}
\; dw\; dz.$$
Note that while there is no restriction on $k$ the sum only contains a
finite number of terms. Continuing,
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon} 
\frac{(1+z)^{p+n}}{z^{n+1}} 
\frac{1}{2\pi i}
\int_{|w|=\gamma} 
\frac{(1+w)^{q+n}}{w^{n+1}} 
\\ \times
\left(1-\frac{z w}{(1+z)(1+w)}\right)^{p+q+1}
\; dw\; dz$$
or $$\frac{1}{2\pi i}
\int_{|z|=\epsilon} 
\frac{(1+z)^{n-q-1}}{z^{n+1}} 
\frac{1}{2\pi i}
\int_{|w|=\gamma} 
\frac{(1+w)^{n-p-1}}{w^{n+1}} 
(1+ z + w)^{p+q+1}
\; dw\; dz$$
Supposing that $p\ge n$ and $q\ge n$
and $\epsilon \ll 1$ and $\gamma \ll 1$  this may be re-written as
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon} 
\frac{1}{z^{n+1} (1+z)^{q+1-n}} 
\frac{1}{2\pi i}
\int_{|w|=\gamma} 
\frac{1}{w^{n+1} (1+w)^{p+1-n}} 
\\ \times (1+ z + w)^{p+q+1}
\; dw\; dz$$
Put $w = (1+z) u$ so that $dw = (1+z) \; du$ to get
with $\delta \lt \gamma/(1+\epsilon)$
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon} 
\frac{1}{z^{n+1} (1+z)^{q+1-n}} 
\frac{1}{2\pi i}
\int_{|u|=\delta} 
\frac{1}{(1+z)^{n+1} u^{n+1} (1+(1+z)u)^{p+1-n}} 
\\ \times (1+ z)^{p+q+1} (1+u)^{p+q+1}
  \; (1+z) \; du\; dz$$
Note that the pole at $u=-1/(1+z)$ has norm $\delta/\gamma \gt \delta$
so it is not inside the contour in $u$. This yields
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon} 
\frac{(1+z)^p}{z^{n+1}} 
\frac{1}{2\pi i}
\int_{|u|=\delta} 
\frac{1}{u^{n+1} (1+ u + z u)^{p+1-n}} 
\\ \times (1+u)^{p+q+1}
\; du\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} 
\frac{(1+z)^p}{z^{n+1}} 
\frac{1}{2\pi i}
\int_{|u|=\delta} 
\frac{(1+u)^{n+q}}{u^{n+1} (1 + z u /(1+u))^{p+1-n}} 
\; du\; dz$$
Extracting the residue for $z$ first we obtain
$$\sum_{k=0}^n {p\choose n-k}
\frac{(1+u)^{n+q}}{u^{n+1}}
{k+p-n\choose k}
(-1)^k \frac{u^k}{(1+u)^k}.$$
The residue for $u$ then yields $$\sum_{k=0}^n (-1)^k {p\choose n-k}
{k+p-n\choose k} {n-k+q\choose n-k}.$$
The sum term here is $$\frac{p!\times (p+k-n)!\times (q+n-k)!}
{(n-k)! (p+k-n)! \times k! (p-n)! \times (n-k)! q!}$$ which simplifies
to $$\frac{p!\times n! \times (q+n-k)!}
{(n-k)! \times n!\times  k! (p-n)! \times (n-k)! q!}$$ which is
$${n\choose k} {p\choose n}{q+n-k\choose q}$$ so we have for the sum
$${p\choose n}
\sum_{k=0}^n {n\choose k} (-1)^k {q+n-k\choose q}.$$
To evaluae the remaining sum we introduce $${q+n-k\choose q}
= \frac{1}{2\pi i}
\int_{|v|=\epsilon} \frac{(1+v)^{q+n-k}}{v^{q+1}} \; dv$$
getting for the sum $${p\choose n}
\frac{1}{2\pi i}
\int_{|v|=\epsilon}
\frac{(1+v)^{q+n}}{v^{q+1}}
\sum_{k=0}^n {n\choose k} (-1)^k \frac{1}{(1+v)^k} \; dv
\\ = {p\choose n}
\frac{1}{2\pi i}
\int_{|v|=\epsilon}
\frac{(1+v)^{q+n}}{v^{q+1}}
\left(1-\frac{1}{1+v}\right)^n \; dv
\\ = {p\choose n}
\frac{1}{2\pi i}
\int_{|v|=\epsilon}
\frac{(1+v)^{q}}{v^{q-n+1}} \; dv
= {p\choose n} {q\choose q-n}$$
which is $${p\choose n} {q\choose n}.$$
This concludes the argument.
