Derivation of formula involving Gamma function? I'm trying to prove that:
$$\prod_{n=1}^{\infty}\frac{n(n+a+b)}{(n+a)(n+b)} = \frac{\Gamma(a+1)\Gamma(b+1)}{\Gamma(a+b+1)}$$
whenever $a$ and $b$ are positive.
I know that
$$\frac{\Gamma(a+1)\Gamma(b+1)}{\Gamma(a+b+1)} = \frac{\int_0^{\infty}e^{-s}s^ads \int_0^{\infty}e^{-t}t^adt}{\int_0^{\infty}e^{-n}n^{a+b}dn}$$
but am confused as to where to proceed from here... Should I use the product formula for $1/\Gamma$ instead? Any direction would be appreciated. Thanks.
 A: Another hint (bit more rigorous perhaps): The Beta Function can be written as
\begin{equation}
B(x,y) = \frac{x+y}{x y} \prod_{n=1}^\infty \left( 1+ \dfrac{x y}{n (x+y+n)}\right)^{-1},
\end{equation}
the right hand side of the above formula can be expanded
\begin{eqnarray}
B(x,y) &=& \frac{x+y}{x y} \prod_{n=1}^\infty \left( 1+ \dfrac{x y}{n (x+y+n)}\right)^{-1} \\
       &=& \frac{x+y}{x y} \prod_{n=1}^\infty \left( \frac{n(x+y+n)+xy}{n(x+y+n)} \right)^{-1} \\
       &=& \frac{x+y}{x y} \prod_{n=1}^\infty \left( \frac{n(x+y+n)}{xy+n(x+y)+n^{2}}\right) \\
       &=& \frac{x+y}{x y} \prod_{n=1}^\infty \left( \frac{n(x+y+n)}{(n+x)(n+y)}\right)
\end{eqnarray}
Now; The Beta function has many other forms such as
\begin{equation}
B(x,y)=\dfrac{\Gamma(x)\,\Gamma(y)}{\Gamma(x+y)}
\end{equation}
From which a relationship between your left hand side and your right hand side could become readily found once you use properties of the gamma fuction.
A: Hint: Use Euler's infinite product formula for the $\Gamma$ function.
A: First note that $x^{(n)}$, denotes the rising factorial and
$$ x^{(n)}=x(x+1)(x+2)\cdots(x+n-1)=\frac{\Gamma(x+n)}{\Gamma(x)} $$
So now
$$
\prod_{n=1}^{\infty}\frac{n(n+a+b)}{(n+a)(n+b)} $$
$$=\lim_{z\to\infty} \prod_{n=1}^{z}\frac{n(n+a+b)}{(n+a)(n+b)}
$$
$$
= \lim_{z\to\infty} \frac{\left[\prod\limits_{n=1}^{z} n\right]\left[\prod\limits_{n=1}^{z} (n+a+b)\right]}{\left[\prod\limits_{n=1}^{z} (n+a)\right]\left[\prod\limits_{n=1}^{z} (n+b)\right]}$$
$$= \lim_{z\to\infty} \frac{z!(a+b+1)^{(z)}}{(a+1)^{(z)}(b+1)^{(z)}}
$$
$$
= \lim_{z\to\infty} \frac{\Gamma(z+1)\frac{\Gamma (a+b+1+z)}{\Gamma(a+b+1)}}{\left(\frac{\Gamma(a+1+z)}{\Gamma(a+1)}\right)\left(\frac{\Gamma(b+1+z)}{\Gamma(b+1)}\right)}$$
$$=\lim_{z\to\infty} \frac{\Gamma(z+1)\Gamma (a+b+1+z)\Gamma(a+1)\Gamma(b+1)}{\Gamma(a+1+z)\Gamma(b+1+z)\Gamma(a+b+1)}
$$
$$
=\frac{\Gamma(a+1)\Gamma(b+1)}{\Gamma(a+b+1)}\lim_{z\to\infty} \frac{\Gamma(z+1)\Gamma (a+b+1+z)}{\Gamma(a+1+z)\Gamma(b+1+z)}$$
$$=\frac{\Gamma(a+1)\Gamma(b+1)}{\Gamma(a+b+1)}
$$
