TITLE: How to find $f(m)=\prod\limits_{n=2}^{\infty}\left(1-\frac{1}{n^m}\right)^{-1}$ (if $m>1$)?
QUESTION [5 upvotes]: 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?

REPLY [4 votes]: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \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
  $\ds{\sum_{k = 1}^{m}\omega_{k} =
\sum_{k = 1}^{m}\exp\pars{2\pi k\ic \over m} = \color{red}{0}}$.