Infinite Product computation How can we compute the infinite product:

Do we need Gamma function?
Edited: I forgot to add 1/e factor so the product converges.
The product becomes

 A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
 \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack}
 \newcommand{\dd}{{\rm d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}}
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{{\rm i}}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\Li}[1]{\,{\rm Li}_{#1}}
 \newcommand{\pars}[1]{\left(\, #1 \,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,}
 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$
Note that
$$
\prod_{k=2}^{\infty}\bracks{\frac{1}{\expo{}}\pars{1 + \frac{1}{k^{2} - 1}}
^{k^{2}\ - 1}}
=\exp\pars{\color{#66f}{%
\sum_{k=2}^{\infty}
\braces{\bracks{k^{2} - 1}\ln\pars{1 + \frac{1}{k^{2} - 1}} - 1}}}
$$
We'll study the above $\ds{\color{#66f}{blue}}$ expression:
\begin{align}
&\color{#66f}{\lim_{N \to \infty}\
\sum_{k=2}^{N}\bracks{\pars{k^{2} - 1}\ln\pars{1 + \frac{1}{k^{2} - 1}}- 1}}
=\lim_{N \to \infty}\ \sum_{k=2}^{N}\bracks{\pars{k^{2} - 1}\ \overbrace{%
2\int_{0}^{1}\frac{x\,\dd x}{k^{2} - x^{2}}}
^{\dsc{\ln\pars{1 + \frac{1}{k^{2} - 1}}}}\ -\ 1}
\\[5mm]&=\lim_{N \to \infty}\ \sum_{k=2}^{N}\bracks{\pars{k^{2} - 1}
2\int_{0}^{1}\frac{x\,\dd x}{k^{2} - x^{2}}\ -\
\overbrace{2\int_{0}^{1}x\,\dd x}^{\dsc{1}}}
\\[5mm]&=2\lim_{N \to \infty}\ \sum_{k=2}^{N}\
\int_{0}^{1}\frac{\pars{k^{2}x - x}-\pars{k^{2}x - x^{3}} }{k^{2} - x^{2}}\,\dd x
\\[5mm]&=2\int_{0}^{1}x\pars{x^{2} - 1}
\sum_{k=0}^{\infty}\frac{1}{\pars{k + 2 + x}\pars{k + 2 - x}}\,\dd x
\\[5mm]&=\int_{0}^{1}
\pars{x^{2} - 1}\bracks{\Psi\pars{2 + x} - \Psi\pars{2 - x}}\,\dd x\tag{1}
\end{align}
where $\ds{\Psi}$ is the Digamma Function. With the Digamma Recurrence Formula:
\begin{align}
\Psi\pars{2 + x} - \Psi\pars{2 - x}
&=\bracks{\Psi\pars{1 + x} + \frac{1}{1 + x}}
-\bracks{\Psi\pars{1 - x} + \frac{1}{1 - x}}
\\[5mm]&=\frac{1}{1 + x} - \frac{1}{1 - x} + \Psi\pars{1 + x}
-\bracks{\Psi\pars{-x} + \frac{1}{-x}}
\\[5mm]&=\frac{1}{1 + x} - \frac{1}{1 - x} + \frac{1}{x}
+ \bracks{\Psi\pars{1 + x} - \Psi\pars{-x}}
\\[5mm]&=\frac{1}{1 + x} - \frac{1}{1 - x} + \frac{1}{x} - \pi\cot\pars{\pi x}
\tag{2}
\\[5mm]&=\totald{}{x}\bracks{\ln\pars{1 + x} + \ln\pars{1 - x} + \ln\pars{x}\
-\ln\pars{\sin\pars{\pi x}}}
\end{align}
where we used, in expression $\pars{2}$, the Euler Reflection Formula. Expression $\pars{1}$ is reduced to:
\begin{align}
&\color{#66f}{\lim_{N \to \infty}\
\sum_{k=2}^{N}\bracks{\pars{k^{2} - 1}\ln\pars{1 + \frac{1}{k^{2} - 1}}- 1}}
\\[5mm]&=-\ln\pars{\pi}
-2\ \underbrace{%
\int_{0}^{1}\bracks{\ln\pars{1 + x} + \ln\pars{1 - x} + \ln\pars{x}\
-\ln\pars{\sin\pars{\pi x}}}x\,\dd x}
_{\dsc{\frac{2\ln\pars{2} - 3}{4}}}
\end{align}

Finally,
$$
\color{#66f}{\large\prod_{k=2}^{\infty}\bracks{\frac{1}{\expo{}}\pars{1 + \frac{1}{k^{2} - 1}}^{k^{2}\ - 1}}}
=
\color{#66f}{\large\frac{\expo{3/2}}{2\pi}}\approx{\tt 0.7133}
$$
