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

$$\frac{\sin(\pi z)}{\pi z}=\prod_{n\geq 1}\left(1-\frac{z^2}{n^2}\right),\qquad \frac{\sinh(\pi z)}{\pi z}=\prod_{n\geq 1}\left(1+\frac{z^2}{n^2}\right)\tag{1}$$ 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.