A Partial product involving the Gamma function I previously asked this question about the shape of the following infinite product involving the Gamma function.  
$$\prod_{n=1}^\infty\frac{\left(\Gamma(n+1)\right)^2}{\Gamma\left(n+x+1\right)\Gamma\left(n-x+1\right)}=\prod_{n=1}^\infty\prod_{k=1}^\infty\left(1-\frac{x^2}{\left(n+k\right)^2}\right)$$
It appeared to fit a normal curve very well for more terms, which made me wonder if it had the form of an exponential.  The problem, as you can see from @marty cohen’s answer, is that the product diverges.
But what about the partial product?
$$\prod_{n=1}^m\frac{\left(\Gamma(n+1)\right)^2}{\Gamma\left(n+x+1\right)\Gamma\left(n-x+1\right)}$$
Can we find an exact or asymptotic formula to shed some light on the shape of this function?
 A: Revised version after @tyobrien key remark.
As I wrote in my first answer to your previous question, there is a closed form expressionfor$$f_p(x)=\prod_{n=1}^p\frac{\left(\Gamma(n+1)\right)^2}{\Gamma\left(n+x+1\right)\,\Gamma\left(n-x+1\right)}$$
It is given in terms of  the Barnes G-function (sometimes named  the double gamma function) and write
$$f_p(x)=\frac{ G(p+2)^2 \, G(2-x)\, G(2+x) }{G(p+2-x)\, G(p+2+x)}$$
This $G(.)$ is pretty documented (google for it) and seems to be related to many other functions (have a look here). 
Concerning the asymptotics, as I wrote in my second answer to your previous question, it seems that, building the Taylor expansion around $x=0$, we have very good approximations looking like
$$f_p(x)=\exp\left(-\sum_{i=1}^\infty c_k^{(p)} x^{2k} \right)$$ where all coefficients have an explicit formulation in terms of polygamma and zeta functions (all of them are positive). To make them clearer and to see some possible patters, writing
$$c_k^{(p)}=d_k^{(p)}+e_k^{(p)}$$ the first are given in the table below
$$\left(
\begin{array}{ccc}
k & d_k^{(p)} & e_k^{(p)} \\
 1 & \psi ^{(0)}(p+2)+(p+1) \psi ^{(1)}(p+2) & -\frac{\pi ^2}{6}+\gamma \\
 2 & \frac{3 \psi ^{(2)}(p+2)+(p+1) \psi ^{(3)}(p+2)}{12}  & -\frac{\pi
   ^4}{180}+\frac{\zeta (3)}{2} \\
 3 & \frac{5 \psi ^{(4)}(p+2)+(p+1) \psi ^{(5)}(p+2)}{360}  & -\frac{\pi
   ^6}{2835}+\frac{\zeta (5)}{3} \\
 4 & \frac{7 \psi ^{(6)}(p+2)+(p+1) \psi ^{(7)}(p+2)}{20160} & -\frac{\pi
   ^8}{37800}+\frac{\zeta (7)}{4} \\
 5 & \frac{297 \psi ^{(8)}(p+2)+33 (p+1) \psi ^{(9)}(p+2)}{59875200} & -\frac{\pi
   ^{10}}{467775}+\frac{\zeta (9)}{5} \\
 6 & \frac{11 \psi ^{(10)}(p+2)+(p+1) \psi ^{(11)}(p+2)}{239500800} & -\frac{691 \pi^{12}}{3831077250}+\frac{\zeta (11)}{6}\\
 7 & \frac{39 \psi ^{(12)}(p+2)+3 (p+1) \psi ^{(13)}(p+2)}{130767436800} &-\frac{2 \pi ^{14}}{127702575}+\frac{\zeta (13)}{7}
\end{array}
\right)$$
This does confirm your interesting observation.
By the way, you could be interested by this paper.
Numerical aspects
As said earlier, the first term is, from far away, the most significant. In order to check, I computed
$$\Phi(a)=\int_{-3}^3 \left(f_{100}(x)-e^{-a x^2}\right)^2\,dx$$ which I minimized with respect to $a$. The optimum is found for $a=4.5645$ to be compared with 
$c_1^{(100)}=4.5474$.
A: This is too long for a comment.
Let us consider
$$a_p=\left(\frac{\sin(\pi x)}{(\pi x) \prod_{n=1}^{p+1}\left(1-\left(\frac{x}{n}\right)^2\right)}\right)^{p+1}$$
$$\prod_{n=1}^{p+1}\left(1-\left(\frac{x}{n}\right)^2\right)=\frac{(1-x)_{p+1} (x+1)_{p+1}}{((p+1)!)^2}$$
$$\log(a_p)=(p+1) \log \left(\frac{((p+1)!)^2 \sin (\pi  x)}{(\pi  x) (1-x)_{p+1} (x+1)_{p+1}}\right)$$ Now, expand as Taylor series around $x=0$ to get
$$\log(a_p)=-(p+1) \psi ^{(1)}(p+2)x^2+O\left(x^4\right)$$ Now, using asymtotics
$$-(p+1) \psi ^{(1)}(p+2)=-1+\frac{1}{2 p}-\frac{2}{3 p^2}+O\left(\frac{1}{p^3}\right)$$
