Asymptotic formula for $\prod_{k=1}^{\infty}\zeta (2kn)$ Suppose $n\geq 1$ is a positive integer. Can we find an asymptotic formula for this product below.
$$\prod_{k=1}^{\infty}\zeta (2kn)=\zeta (2n)\zeta (4n)\zeta (6n) \cdots$$
I tried to use $\zeta (2n)=\dfrac{(-1)^{n+1}B_{2n}(2\pi )^{2n}}{2(2n)!}$ but couldn't get anywhere.
 A: Let's examine the most significant terms of the product for $n\gg 1$ :
\begin{align}
\tag{1}P_n&:=\prod_{k=1}^{\infty}\zeta (2kn)\\
&=\zeta (2n)\;\zeta (4n)\cdots\\
\tag{*}&=(1+2^{-2n}+3^{-2n}+4^{-2n}+o(4^{-2n}))\;(1+2^{-4n}+o(4^{-2n}))\;(1+o(4^{-2n}))\\
\tag{2}P_n&=1+2^{-2n}+3^{-2n}+2\cdot 4^{-2n}+o(4^{-2n})\\
\end{align}
(the product of the remaining terms $\,\zeta(6n)\,\zeta(8n)\cdots$ in $(*)$ is rewritten $(1+o(4^{-2n}))$ since the most significant term (except $1$) is $\;2^{-6n}+2^{-8n}+\cdots=\dfrac {4^{-2n}}{4^n-1}$)
From $(2)$ we deduce the simple :
$\boxed{\displaystyle P_n-1\sim 4^{-n}}\;$ which could be obtained with Igor Rivin's hint
(the coefficient $-1.390256$ in Claude's approximation is near $-\log(4)$, the constant term should disappear for large $n$...)
Concerning $P_n$ for small values of $n$ : $P_1=C_2$ was considered in :


*

*Bernd Kellner : "On asymptotic constants related to products of Bernoulli numbers and factorials" and at bernoulli.org
($C_2$ is written as an infinite product of Dedekind 
eta function there) ; concerning $C_1$ we have $\;C_1=\lim_{n\overset{>}{\to}\frac 12} (2n-1)P_n\;$ with $P_n$ becoming infinite as $n{\to}\frac 12$) 

*Steven Finch : "Minkowski-Siegel Mass Constants" may also be of interest

A: For a reasonably large $n$ (that is, bigger than about $3$), $\zeta(n) \approx 1+ \frac{1}{2^n},$ which will give you the asymptotics you crave.
A: This is not an answer since just based on numerical simulation.
Being stuck with any formal approach I tried, I just used numerical simulation and what I observed is that, if $$P_n=\prod_{k=1}^{\infty}\zeta (2kn)$$ then $\log(P_n-1)$ varies as a linear function of $n$. A basic linear regression $(1\leq n\leq 50)$ gave $$\log(P_n-1)=0.135417-1.390256 n$$ both parameters being statistically significant.
Edit
Inspired by Igor Rivin's and Raymond Manzoni's answers.
Limiting the $\zeta$ function to the first term $(\zeta(s)=1+\frac 1 {2^s})$, it seems that, for large $n$, we can approximate $$P_n\approx 1+\frac 1 {4^n}$$ since $$\log(P_n)=\sum_{k=1}^\infty \log(\zeta(2kn))\approx\sum_{k=1}^\infty \log(1+\frac 1{2^{2kn}})\approx\sum_{k=1}^\infty \frac 1{2^{2kn}}=\frac{1}{4^n-1}$$ $$P_n\approx \exp(\frac{1}{4^n-1})\approx 1+\frac{1}{4^n}$$ $$\log(P_n-1)\approx -n \log(4)$$ ($\log(4)\approx 1.38629$ being quite close to the $1.390256$ from the empirical calculation).
A: It's senseful to use the Euler product formula for the Riemann zeta function.
With $\displaystyle \zeta(s)=\prod\limits_{p\, prime}\frac{1}{1-p^{-s}}$ and $\displaystyle -\ln (1-x)=\sum\limits_{k=1}^\infty\frac{x^k}{k}$ with $|x|<1$ one gets 
$$\prod\limits_{k=1}^\infty\zeta(2nk)=\exp\sum\limits_{p\, prime}\sum\limits_{k=1}^\infty\frac{1}{k(p^{2nk}-1)}$$
The convergence is good (because of $p^{2nk}$). 
And it becomes clear: $\,\lim\limits_{n\to\infty}\prod\limits_{k=1}^\infty\zeta(2nk)=1$ 
