Prove that $\prod_{n=2}^\infty \frac{1}{e^2}\left(\frac{n+1}{n-1}\right)^n=\frac{e^3}{4\pi}$

The logarithm of the LHS equals

$$ L=\sum_{n\geq 2}\left[-2+2n\,\text{arctanh}\frac{1}{n}\right]=\sum_{n\geq 2}\sum_{m\geq 1}\frac{1}{\left(m+\frac{1}{2}\right)n^{2m}}=\sum_{m\geq 1}\frac{\zeta(2m)-1}{m+\frac{1}{2}}\tag{1} $$ and we may recall that $$ \frac{1-\pi z\cot(\pi z)}{2} = \sum_{m\geq 1}\zeta(2m) z^{2m} \tag{2} $$ to get that: $$ L = \int_{0}^{1}\left(\frac{1-3z^2}{1-z^2}-\pi z\cot(\pi z)\right)\,dz\tag{3} $$ and $L=3-\log(4\pi)$ follows from the connection between the primitive of $z\cot(z)$ and the dilogarithm, proving the stated conjecture: $$ \int z\cot(z)\,dz = z\log\left(1-e^{2iz}\right)-\frac{i}{2}\left(z^2+\text{Li}_2(e^{2iz})\right)\tag{4} $$

Nice question!

Consider the partial product

$$P_N = \prod_{n=2}^{N} \frac1{e^2} \left (\frac{n+1}{n-1} \right )^n $$

By writing out the terms of $P_N$, we can easily find a simple expression for it:

$$P_N = e^{-2(N-1)} \frac{N^{N-1} (N+1)^N}{2 (N-1)!^2} $$

The result follows by using Stirling's approximation as well as the definition of $e$ as $N \to \infty$:

$$(N-1)! \approx \sqrt{2 \pi (N-1)} (N-1)^{N-1} e^{-(N-1)} $$

$$ (N+1)^N \approx e N^n $$

First, recombining terms gives $$ \begin{align} \prod_{n=2}^m\color{#C00}{\frac{1}{e^2}}\left(\frac{\color{#090}{n+1}}{\color{#00F}{n-1}}\right)^n &=\color{#C00}{\frac1{e^{2(m-1)}}}\color{#090}{\prod_{n=3}^{m+1}n^{n-1}}\color{#00F}{\prod_{n=1}^{m-1}n^{-n-1}}\\ &=\frac1{e^{2(m-1)}}\left(\frac{(m+1)^m}2\prod_{n=1}^mn^{n-1}\right)\left(m^{m+1}\prod_{n=1}^mn^{-n-1}\right)\\ &=\frac1{e^{2(m-1)}}\frac{(m+1)^mm^{m+1}}{2m!^2} \end{align} $$ Then, applying Stirling, we get $$ \begin{align} \lim_{m\to\infty}\frac1{e^{2(m-1)}}\frac{(m+1)^mm^{m+1}}{2m!^2} &=\lim_{m\to\infty}\frac1{e^{2m-3}}\frac{m^{2m+1}e^{2m}}{4\pi m^{2m+1}}\\ &=\frac{e^3}{4\pi} \end{align} $$