How to find $f(m)=\prod\limits_{n=2}^{\infty}\left(1-\frac{1}{n^m}\right)^{-1}$ (if $m>1$)?

Hint. By recalling the Weierstrass infinite product of the gamma function, $$ \Gamma(1+z) = e^{-\gamma z} \prod_{n=1}^\infty \left(1+ \frac z n \right)^{-1} e^{z/n},\qquad \text{Re}z>-1, \tag1 $$ then, by writing each factor of the given product over the complex numbers using roots of unity, $$ \left(1-\frac{1}{n^m}\right)^{-1}=\prod_{k=1}^{m}\left(1-\frac{e^{2ik\pi/m}}{n}\right)^{-1},\qquad n\ge2, \tag2 $$ one gets

$$ f(m)=\prod\limits_{n=2}^{\infty}\left(1-\frac{1}{n^m}\right)^{-1}=\prod_{k=1}^{m}\Gamma\left(2-e^{2ik\pi/m}\right) \tag3 $$

from which many particular cases are deduced.

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With $\ds{N \in \mathbb{N}_{\ \geq\ 2}}$ and $\ds{\omega_{k} = \exp\pars{2k\pi\ic/m}}$:

\begin{align} &\bbox[10px,#ffd]{\ds{\left.\prod_{n = 2}^{N}\pars{1 - {1 \over n^{m}}}^{-1}\,\right\vert_{\ m\ \in\ \mathbb{N}_{\ \geq\ 2}}}} = \prod_{n = 2}^{N}{n^{m} \over n^{m} - 1} \\[5mm] = &\ \pars{N!}^{m}\prod_{n = 2}^{N} {1 \over \pars{n - \omega_{1}}\cdots \pars{n - \omega_{m}}} \\[5mm] = &\ \pars{N!}^{m}\prod_{k = 1}^{m}{1 \over \pars{2 - \omega_{k}}^{\overline{N - 1}}} \\[5mm] = &\ \pars{N!}^{m}\prod_{k = 1}^{m}{\Gamma\pars{2 - \omega_{k}} \over \Gamma\pars{N + 1 - \omega_{k}}} \\[5mm] \stackrel{\mrm{as}\ N\ \to\ \infty}{\sim}\,\,\,& \bracks{\prod_{k = 1}^{m}\Gamma\pars{2 - \omega_{k}}} \prod_{k = 1}^{m}{\root{2\pi}N^{N + 1/2}\expo{-N} \over \root{2\pi}\pars{N - \omega_{k}}^{N - \omega_{k} + 1/2} \expo{-\pars{N -\omega_{k}}}} \\[5mm] = & \bracks{\prod_{k = 1}^{m}\Gamma\pars{2 - \omega_{k}}} \prod_{k = 1}^{m}\bracks{{N^{N + 1/2} \over N^{N - \omega_{k} + 1/2}} \,{\expo{-\omega_{k}} \over \pars{1 - \omega_{k}/N}^{N - \omega_{k} + 1/2}}} \\[5mm] \stackrel{\mrm{as}\ N\ \to\ \infty}{\sim}\,\,\, & \bracks{\prod_{k = 1}^{m}\Gamma\pars{2 - \omega_{k}}} \prod_{k = 1}^{m}N^{\omega_{k}} = \bracks{\prod_{k = 1}^{m}\Gamma\pars{2 - \omega_{k}}} N^{\sum_{k = 1}^{m}\omega_{k}} \\[5mm] \stackrel{\mrm{as}\ N\ \to\ \infty}{\to} &\ \bbx{\prod_{k = 1}^{m}\Gamma\pars{2 - \exp\pars{2\pi k\ic \over m}}} \\[2mm] &\ \pars{\begin{array}{l} \mbox{Note that}\ m = 2\ yields\ \Gamma\pars{3}\Gamma\pars{1} = \color{red}{2} \\ \mbox{which is the OP particular example.} \\[3mm] \mbox{In addition, it agrees with the particular} \\ \mbox{values of}\ \color{#66f}{\texttt{@Claude Leibovici}}\ \mbox{answer.} \end{array}} \end{align}

Note that $\ds{\sum_{k = 1}^{m}\omega_{k} = \sum_{k = 1}^{m}\exp\pars{2\pi k\ic \over m} = \color{red}{0}}$.

If you look here and rework a little bit the formulae for $\sinh(x)$ and $\cosh(x)$, you could find that, if $$p_m=\prod\limits_{n=2}^{\infty}\left(1-\frac{1}{n^m}\right)^{-1}$$ $$p_3= 3\pi \, \text{sech}\left(\frac{\sqrt{3} \pi }{2}\right)$$ $$p_4=4 \pi\, \text{csch}(\pi )$$ For $m>4$, I imagine that this would be given by some ugly gamma functions.

Using a CAS, I found that $$p_6=6 \pi ^2 \,\text{sech}^2\left(\frac{\sqrt{3} \pi }{2}\right)$$