MoebiusMu product Some time back, I asked this question about 
$$\prod _{n=1}^{\infty } n^{\mu (n)}.$$
Recently, I found this numeric convergence: $0.6784223987077668596042536007\dots$, and when rationalized, found this:  $\frac{5040}{7429}$.
Factoring the fraction: $\lbrace\lbrace2, 4\rbrace, \lbrace3, 2\rbrace, \lbrace5, 1\rbrace, \lbrace7, 1\rbrace, \lbrace17, -1\rbrace, \lbrace19, -1\rbrace, \lbrace23, -1\rbrace\rbrace$, where $\lbrace prime, exponent\rbrace$, we get this sequence.
I like the fact that the sequence is finite and am wondering what it is doing in my fraction.
Is this coincidence? Or, what?

The product is clearly divergent (the Möbius function takes the values $-1,\,0,\,1$ and we will multiply or divide by $n$ for an infinite number of $n$) but we may observe that :
$$\tag{1}\log\left(\prod _{n=1}^{\infty } n^{\mu (n)}\right)=\sum_{n=1}^\infty \mu (n)\log(n)$$ 
and it is well known that (at least for $\Re(s)>1$) we have : 
$$\tag{2}\sum_{n=1}^\infty \frac{\mu(n)}{n^s}=\frac 1{\zeta(s)}$$
This is true too for $s=1$ but according to G. Hardy the proof of this (given by Landau I think) is equivalent to the PNT.
From this we may deduce, using $n^{-s}=e^{-s\log(n)}$ and supposing that $(2)$ may be extended for $s\in [0,1)$, that :
$$\sum_{n=1}^\infty \mu (n)\log(n)=-\lim_{s\to 0^+} \frac d{ds}\sum_{n=1}^\infty \mu(n)\,e^{-s\log(n)}=\frac {\zeta'(0)}{\zeta(0)^2}=\frac{-\log(2\pi)/2}{(-1/2)^2}=-2\,\log(2\pi)$$
so that, if an answer has to be provided, I would suggest :
$$\prod _{n=1}^{\infty } n^{\mu (n)}=\frac 1{4\pi^2}$$
The method used here is named zeta function regularization and was found pretty efficient to get actual values in physics (especially in QFT).
(Hmmm I notice now, following your link, that the same answer was provided in MO ; sorry for the repetition...)
