Prove an equality between infinite product and gamma function Is it true that : 
$ \forall (x,y) \in \mathbb{R}^2 $
$$ 
\prod_{n=1}^{+\infty} \frac{(n+x)(n+y)}{n(n+x+y)} = \frac{\Gamma(x+y+1)}{\Gamma(x+1)\Gamma(y+1)} \;?
$$
Thank's for your answer ! :) 
 A: Using the functional equation $s\Gamma(s)=\Gamma(s+1)$, which is true whenever $s \notin \{0,-1,-2,\dots\}$, we can write the term of the product as
$$a_n=\frac{\Gamma(n+x+1)}{\Gamma(n+x)}\cdot\frac{\Gamma(n+y+1)}{\Gamma(n+y)}\cdot\frac{\Gamma(n)}{\Gamma(n+1)}\cdot\frac{\Gamma(n+x+y)}{\Gamma(n+x+y+1)}$$
Now, looking at the sequence of finite products $p_k=\prod_{n=1}^ka_n$, we see it is a 'telescoping product' and since $\Gamma(1)=1$ we have that
\begin{aligned}
p_k&=\frac{\Gamma(x+k+1)}{\Gamma(x+1)}\cdot\frac{\Gamma(y+k+1)}{\Gamma(y+1)}\cdot\frac{\Gamma(1)}{\Gamma(k+1)}\cdot\frac{\Gamma(x+y+1)}{\Gamma(x+y+k+1)}\\
&=\frac{\Gamma(x+k+1)\cdot\Gamma(y+k+1)}{\Gamma(k+1)\cdot\Gamma(x+y+k+1)}\cdot\frac{\Gamma(x+y+1)}{\Gamma(x+1)\cdot\Gamma(y+1)}
\end{aligned}
Thus, we need only show that \begin{equation}\label{eq1}\tag{*}\lim\limits_{k\to\infty}\frac{\Gamma(x+k+1)\cdot\Gamma(y+k+1)}{\Gamma(k+1)\cdot\Gamma(x+y+k+1)}=1\end{equation}
To that end, we shall use the following asymptotic estimate for $\Gamma$:

Lemma: For all $\alpha \in \mathbb{R}$ it holds that:
$$\lim\limits_{n\to \infty}\frac{\Gamma(n+\alpha)}{\Gamma(n)\cdot n^{\alpha}}=1$$

Now, rewrite the expression in \eqref{eq1} to:
\begin{equation}
\frac{\Gamma(x+k+1)}{\Gamma(k+1)\cdot (k+1)^x}\cdot\frac{\Gamma(y+k+1)}{\Gamma(k+1)\cdot (k+1)^y}\cdot\frac{\Gamma(k+1)\cdot (k+1)^{x+y}}{\Gamma(x+y+k+1)}
\end{equation}
and simply apply the lemma to each term in the product.
Thus, the answer is mostly yes. When we employed the functional equation for $\Gamma$, we implicitly assumed $n+x$, $n+y$ and $n+x+y$ not to be in $\{0,-1,-2,\dots\}$, that is, $x$ and $y$ are assumed not to be negative integers.
When one or more of them are, you can do some analysis on cancellation of poles with zeroes (remember $\Gamma$ has no zeroes, simple poles at non-positive integers and no poles elsewhere), but I'm not sure how much you're interested in this particularity and in any case you still have to specify how to make sense of either side of the equation.
