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: equation number (10) in the link.