<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] 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 $\displaystyle \left(1-\frac{1}{n^4}\right) = \left(1+\frac{1}{n^2}\right) \cdot \left(1-\frac{1}{n^{2}}\right)$
</s>