How do we show that


$$\prod\limits\_{k=1}^{n} \tau(k) = 2^{n (\log \log n + C) + 
\phi(n)},$$


where $\tau(k)$ is the number of divisors of $k$, the constant $C$ is given by 
$$C = \gamma + \sum\_{\nu = 2}^{\infty} \left\{ \log\_2 \left(1 + \frac{1}{\nu}\right) - \frac{1}{\nu} \right\} \left(\sum\limits\_{p \textrm{ prime }} \frac{1}{p^{\nu}} \right),$$
and 
$$\begin{eqnarray\*} 
\frac{\phi(n)}{n} = \frac{\gamma -1}{\log n} + \frac{1!}{(\log n)^2} 
(\gamma + \gamma\_1 -1 ) + \frac{2!}{(\log n)^3} (\gamma + \gamma\_1 + \gamma\_2 
- 1) + \ldots + \frac{(r-1)!}{(\log n)^r} (\gamma + \gamma\_1 + \gamma\_2 + \ldots + 
\gamma\_{r-1} - 1) + O \left\{\frac{1}{(\log n)^{r+1}} \right\}?
\end{eqnarray\*}$$


[Reference:](http://ramanujan.sirinudi.org/Volumes/published/ram17.html#p17_eq1) equation number (10) in the link.



Product of $\tau(k)$
The proof relies on the following identity, which is straightforward to verify:
$$ \sum\_{k=1}^\infty\frac{\log\_2\tau(k)}{k^s}=\zeta(s)\sum\_p\sum\_{\nu=1}^\infty\frac{\log\_2\left(1+\frac{1}{\nu}\right)}{p^{\nu s}},\qquad\Re s>1.$$
Here $p$ runs through the prime numbers. This implies
$$ \sum\_{k=1}^\infty\frac{\log\_2\tau(k)}{k^s}=\zeta(s)\{\log\zeta(s)+g(s)\},\qquad\Re s>1, $$
where
$$g(s):=\sum\_p\sum\_{\nu=2}^\infty\frac{\log\_2\left(1+\frac{1}{\nu}\right)-\frac{1}{\nu}}{p^{\nu s}}$$
is analytic in the half-plane $\Re s>1/2$. From here we can proceed as in Section 3 of [Diaconis (1976)](https://statistics.stanford.edu/sites/default/files/EFS%20NSF%2096.pdf). The result follows upon noting that $C=\gamma+g(1)$.


