How to find $f(m)=\prod\limits_{n=2}^{\infty}\left(1-\frac{1}{n^m}\right)^{-1}$ (if $m>1$)? First case $m=2$ is pretty simple:
$$\prod\limits_{n=2}^{k}\left(1-\frac{1}{n^2}\right)^{-1}=\frac{2k}{k+1}$$
Of course $k>1$ and $\lim\limits_{k\to\infty}f(2)=2$.
How to find $f(m)$ for other cases?
 A: 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.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
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 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
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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
  $\ds{\sum_{k = 1}^{m}\omega_{k} =
\sum_{k = 1}^{m}\exp\pars{2\pi k\ic \over m} = \color{red}{0}}$.

A: 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
$$p_6=6 \pi ^2 \,\text{sech}^2\left(\frac{\sqrt{3} \pi }{2}\right)$$
A: We may also notice that
$$ \prod_{n\geq 2}\left(1-\frac{1}{n^m}\right)^{-1} = \exp\sum_{n\geq 2}-\log\left(1-\frac{1}{n^m}\right)=\exp\sum_{n\geq 2}\sum_{h\geq 1}\frac{1}{h n^{mh}} $$
equals
$$ \exp\left[m\sum_{h\geq 1}\frac{\zeta(mh)-1}{mh}\right]=\exp\left[m\sum_{h\geq 1}\frac{1}{(mh)!}\int_{0}^{+\infty}\frac{x^{mh-1}}{e^{x}-1}\,dx\right] $$
and for integer values of $m\geq 2$ we may deduce a closed form by applying the discrete Fourier transform to the Taylor series of the exponential function, then by invoking the residue theorem.
