Prove that $\prod_{n=2}^∞ \left( 1 - \frac{1}{n^4} \right) = \frac{e^π - e^{-π}}{8π}$ The question
Prove that:
$$\prod_{n=2}^∞ \left( 1 - \frac{1}{n^4} \right) = \frac{e^π - e^{-π}}{8π}$$

What I've tried
Knowing that:
$$\sin(πz) = πz \prod_{n=1}^∞ \left( 1 - \frac{z^2}{n^2} \right)$$
evaluating at $z=i$ gives
$$ \frac{e^π - e^{-π}}{2i} = \sin(πi) = πi \prod_{n=1}^∞ \left( 1 + \frac{1}{n^2} \right)$$
so:
$$ \prod_{n=1}^∞ \left( 1 + \frac{1}{n^2} \right) = \frac{e^π - e^{-π}}{2π}$$
I'm stucked up and don't know how to continue, any help?
 A: $$\frac{\sin(\pi z)}{\pi z}=\prod_{n\geq 1}\left(1-\frac{z^2}{n^2}\right),\qquad \frac{\sinh(\pi z)}{\pi z}=\prod_{n\geq 1}\left(1+\frac{z^2}{n^2}\right)\tag{1}$$
give:
$$ \frac{\sin(\pi z)\sinh(\pi z)}{\pi^2 z^2(1-z^4)}=\prod_{n\geq 2}\left(1-\frac{z^4}{n^4}\right)\tag{2} $$
hence by considering $\lim_{z\to 1}LHS$ we have:
$$ \prod_{n\geq 2}\left(1-\frac{1}{n^4}\right)=\frac{\sinh \pi}{4\pi} = \color{red}{\frac{e^\pi-e^{-\pi}}{8\pi}}\tag{3}$$
as wanted.
A: Note that $$\prod_{n=2}^{\infty} \left(1-\frac{1}{n^{2}}\right) \to \frac{1}{2}$$
This is because $$A_{n} =\prod_{k=2}^{n}\left(1-\frac{1}{n^2}\right) = \prod_{k=2}^{n} \frac{(k-1)(k+1)}{k^2} = \frac{n+1}{2n} \to \frac{1}{2}$$
We have used $\displaystyle \left(1-\frac{1}{n^4}\right) = \left(1+\frac{1}{n^2}\right) \cdot \left(1-\frac{1}{n^{2}}\right)$
A: I'll reproduce the answer @C.Dubussy have just deleted:
$$ \prod_{n=2} \left( 1 - \frac{1}{n^4} \right) = \prod_{n=2}^∞ \left( 1 + \frac{1}{n^2}\right) \prod_{n=2}^∞ \left( 1 - \frac{1}{n^2}\right) = \frac{\sin{iπ}}{iπ} \prod_{n=2}^∞ \frac{n-1}{n}  \prod_{n=2}^∞ \frac{n+1}{n} $$
And because the last product gives $\frac{1}{2}$, we have it!
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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With $\ds{N \in \mathbb{N}_{\ \geq\ 2}}$:

\begin{align}
\prod_{n = 2}^{N}\pars{1 - {1 \over n^{4}}} & =
\prod_{n = 2}^{N}
{\pars{n - 1}\pars{n + 1}\pars{n - \ic}\pars{n + \ic} \over n^{4}}
\\[5mm] & =
{\pars{N - 1}!\bracks{\pars{N + 1}!/2} \over \pars{N!}^{4}}\,
\verts{\pars{2 + \ic}^{\overline{N - 1}}}^{2} =
{1 \over 2}\,{N + 1 \over N}\,\verts{{1 \over N!}\,{\Gamma\pars{N + 1 + \ic} \over \Gamma\pars{2 + i}}}^{\,2}
\\[5mm] & =
{1 \over 2}\,{N + 1 \over N}\,\verts{{1 \over 1 + \ic}
\,{\pars{N + \ic}! \over N!}}^{2}\,{1 \over \verts{\Gamma\pars{1 + i}}^{2}}
\end{align}

With A & S Table $\ds{\mathbf{\color{#000}{6.1.31}}}$ identity
  $\ds{\verts{\Gamma\pars{1 + \ic y}}^{\,2} = {\pi y \over \sinh\pars{\pi y}}}$ and the Stirling Asymptotic Expansion:

\begin{align}
\prod_{n = 2}^{N}\pars{1 - {1 \over n^{4}}} &
\stackrel{\mrm{as}\ N\ \to\ \infty}{\sim}\,\,\,
{1 \over 4}\verts{\root{2\pi}\pars{N + \ic}^{N + \ic + 1/2}\expo{-N - \ic} \over \root{2\pi}N^{N + 1/2}\expo{-N}}^{2}\,{1 \over  \pi/\sinh\pars{\pi}}
\\[5mm] & =
{\expo{\pi} - \expo{-\pi} \over 8\pi}
\,\verts{N^{\ic}\pars{1 + {\ic \over N}}^{N + \ic + 1/2}}^{2} =
{\expo{\pi} - \expo{-\pi} \over 8\pi}
\,\verts{\exp\pars{\ic\ln\pars{N}}\expo{\ic}}^{2}
\\[5mm] & \stackrel{\mrm{as}\ N\ \to\ \infty}{\to}\,\,\,
\bbx{{\expo{\pi} - \expo{-\pi} \over 8\pi}}
\end{align}
