Infinite product of Zetafunctions It is well-known (I learned about this first in a video by Papa Flammy) that $\sum_{n=2}^\infty (\zeta(n)-1) = 1$. This result on its own is quite remarkable, but it also implies convergence of the infinite product
$$P = \prod_{n=2}^\infty \zeta(n).$$
I somehow got invested in this. Upon trying to work out the exact value of $P$, I found that
$$P = \sum_{n=1}^\infty \frac{c_2(n)}n,$$
where $c_2$ is the multiplicative continuation of $c_2(p^e) := b_2(e),$ where $b_2(e)$ is the amount of partitions of $e$ not containing $1$, i.e.
$$b_2(n) = \# \{(a_2, a_3, a_4, \dots ) \in \mathbb N_{\geq 0}^* | 2a_2 + 3a_3 + \dots = n\}.$$
Furthermore, the function
$$Q(s) = \prod_{n=1}^\infty \zeta(ns),$$
converges compactly on $\Re s > 1,$ and one has
$Q(s) = \sum_{n=1}^\infty c_1(n) n^{-s}.$ Here again $c_1(p^e) = b_1(e),$ the amount of partitions of $e$ (this time including 1). Also, $Q$ has Residue
$$\text{Res}_{s=1} (Q(s)) = P.$$
Doing some numerical analysis with python I found that $P \approx 2.29$, but i would love to know if there is some exact form. Does anyone know about such a result?
 A: Yes, for $\Re(s) > 1/2$, $$F(s)=\prod_{n\ge 2} \zeta(ns)= \prod_p \prod_{n\ge 2} \frac1{1-p^{-sn}}=(\prod_p 1+ \sum_{k\ge 1} b_2(k)p^{-sk})=\sum_{m\ge 1} c_2(m)m^{-s}$$
where $b_2(k)$ is the number of partitions in integers $\ge 2$ and $c_2(m)=\prod_{p^k\| m} b_2(k)$.
The product for $F(s)$ converges locally uniformly for $\Re(s)>0, 1/s\not \in \Bbb{Z}_{\ge 2}$, so it is meromorphic for $\Re(s)>0$ with simple poles at $1/s \in \Bbb{Z}_{\ge 2}$.
$Res(F(s),1/N)=\frac1N \prod_{n\ne N} \zeta(n/N)$
$F(s)$ has a naturaly boundary at $\Re(s)=0$ (it cannot be analytically continued further) because the non-trivial zeros of each $\zeta(ns)$ accumulate to the right of $\Re(s)=0$ making $F(s)$ not analytic anywhere on $\Re(s)=0$.
It is mostly obvious that nothing will have a closed-form here.
A: This is practically just a rewriting, and I also doubt there is a closed form.
We have $$\prod_{n\ge 2} \zeta(n) = \prod_p \frac{p-1}{p\,\phi(1/p)}$$
where $\phi(q)$ is the Euler function (the one used in modular forms, not the totient function).
