<s>[INST] 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?
 [/INST] 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!
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