Show that, $\prod_{k=2}^{\infty}\left(\prod_{n=0}^{\infty}(n!)^{\frac{k-1}{k^{n+1}}}\right)=\prod_{m=2}^{\infty}m^{\zeta(s)-1}$ The factorial n! is defined for all positive integer n as
$$n!=n(n-1)(n-2)\cdots2\cdot1$$
The $\zeta(s)$ is defined for all real R(s)>1 as
$$\zeta(s)=\sum_{n=0}^{\infty}\frac{1}{(n+1)^s}$$
Show that,
$$\prod_{k=2}^{\infty}\left(\prod_{n=0}^{\infty}(n!)^{\frac{k-1}{k^{n+1}}}\right)=\prod_{m=2}^{\infty}m^{\zeta(m)-1}$$

Since
$$ \sum_{k\geq 2}\frac{k-1}{k^{n+1}} = \zeta(n)-\zeta(n+1) $$
we have:
$$ \prod_{k\geq 2}\prod_{n\geq 0}n!^{\frac{k-1}{k^{n+1}}} = \prod_{n\geq 2}(n!)^{\zeta(n)-\zeta(n+1)} = \exp\sum_{n\geq 2}\left(\zeta(n)-\zeta(n+1)\right)(\log n!) $$
and the last series equals
$$ \sum_{m\geq 2}\left(\zeta(m)-1\right) \log(m) $$
by summation by parts. The claim easily follows.
