Does this double product equal an exponential function? I was looking at the graph of $$\prod_{n=1}^\infty\frac{\left(\Gamma(n+1)\right)^2}{\Gamma\left(n+x+1\right)\Gamma\left(n-x+1\right)}=\prod_{n=1}^\infty\prod_{k=1}^\infty\left(1-\frac{x^2}{\left(n+k\right)^2}\right)$$
I noticed that it looks much like a normal curve of height $1$.
So is this equal to the form $e^{\frac{-x^2}{v}}$ for some $v$? 
Or essentially the same question: does
$$\sum_{n=1}^\infty\sum_{k=1}^\infty\ln\left(1-\frac{x}{\left(n+k\right)^2}\right)=-\frac{x}{v}$$
for some $v$?
Edit Related Question on the partial product.
 A: A problem is that
the sum diverges.
$\begin{array}\\
s(x)
&=-\sum_{n=1}^\infty\sum_{k=1}^\infty\ln\left(1-\frac{x}{\left(n+k\right)^2}\right)\\
&\ge\sum_{n=1}^\infty\sum_{k=1}^\infty \dfrac{x}{(n+k)^{2}}
\quad\text{since }-\ln(1-z) \ge z \text{ if } z \ge 0\\
&=x\sum_{n=1}^\infty\sum_{k=n+1}^\infty \dfrac{1}{k^{2}}\\
&\ge x\sum_{n=1}^\infty\sum_{k=n+1}^\infty \dfrac{1}{k(k+1)}\\
&= x\sum_{n=1}^\infty\sum_{k=n+1}^\infty (\dfrac1{k}-\dfrac1{k+1})\\
&= x\sum_{n=1}^\infty\dfrac1{n+1}\\
\end{array}
$
and this diverges.
A: I prefer to add another answer for an extension of the work.
Considering
$$f_p(x)=\prod_{n=1}^p\frac{\left(\Gamma(n+1)\right)^2}{\Gamma\left(n+x+1\right)\,\Gamma\left(n-x+1\right)}$$
$$\log\left(f_p(x) \right)=\sum_{n=1}^p \log\left(\frac{\left(\Gamma(n+1)\right)^2}{\Gamma\left(n+x+1\right)\,\Gamma\left(n-x+1\right)} \right)$$
Using Taylor expansion
$$\log\left(\frac{\left(\Gamma(n+1)\right)^2}{\Gamma\left(n+x+1\right)\,\Gamma\left(n-x+1\right)}\right)=-2\sum_{k=1}^\infty \frac{\psi ^{(2 k-1)}(n+1)}{(2 k)!}x^{2k}$$ and what remain is to compute the sums over $n$.
Writing
$$\log\left(f_p(x) \right)=c_1 x^2+c_2 x^4+c_3 x^6 + c_4 x^8+\cdots$$  that is to say
$$f_p(x)=\exp(c_1 x^2+c_2 x^4+c_3 x^6 + c_4 x^8+\cdots)$$
we should have
$$c_1=-\psi ^{(0)}(p+2)-(p+1) \psi ^{(1)}(p+2)-\gamma+\frac{\pi^2}6$$
$$c_2=\frac{-45 \psi ^{(2)}(p+2)-15 (p+1) \psi ^{(3)}(p+2)-90 \zeta
   (3)+\pi ^4}{180} $$
$$c_3=\frac{-315 \psi ^{(4)}(p+2)-63 (p+1) \psi ^{(5)}(p+2)+8 \left(\pi ^6-945 \zeta
   (5)\right)}{22680}$$
$$c_4=\frac{-105 \psi ^{(6)}(p+2)-15 (p+1) \psi ^{(7)}(p+2)+8 \left(\pi ^8-9450 \zeta
   (7)\right)}{302400}$$
For any value of $p$, all coefficients are negative and they are smaller and smaller; this justifies the approximation given by $(3)$ in the previous answer.
For infinitely large values of $p$, we have
$$\frac {c_2}{c_1} \sim \frac{90 \zeta (3)-\pi^4}{30 \left(6 \log (p)-\pi ^2+6 \gamma +6\right)}$$
$$\frac {c_3}{c_2} \sim \frac{4 \left(\pi ^6-945 \zeta (5)\right)}{63 \left(\pi ^4-90 \zeta (3)\right)}\approx 0.109046$$
$$\frac {c_4}{c_3} \sim \frac{3 \left(\pi ^8-9450 \zeta (7)\right)}{40 \left(\pi ^6-945 \zeta (5)\right)}\approx 0.163594$$
Warning
Better material in my answer to this question.
