Questions about Newman's simplified proof of Ramanujan's partition formula

Question

Recently I started to go through Newman's proof of Ramanujan's asymptotic formula for the number of partitions $p(n)$. I got stuck right in the beginning, where we have

$f(z) = \prod_{n=1}^\infty \frac{1}{1-z^n}$

and Newman then uses $\log$ to obtain

$\log(f(z)) = \sum_{n=1}^\infty \frac{1}{n} \frac{z^n}{1-z^n}$.

What's quite clear is that we get some sort of sum using $\log$, e.g. at first

$\log(f(z)) = \sum_{n=1}^\infty \log(\frac{1}{1-z^n}) = \sum_{n=1}^\infty - \log(1-z^n)$.

I just tried my best but failed to prove the postulated identity stated above, which is quite unsatisfying since it's just the introduction to something way more difficult. I would be extremely grateful if someone could help me clear this topic. (This question's already answered, see below.)

EDIT: Since you helped me that fast, I managed to go further into the proof. At first, I was amazed that Newman refers to just three other sources, but I got confused by his reference to Polya/Szegö. I tried to look it up but can't find the right version auf "Lehrsätze" in our library. By now, it's hidden to me what $g$ is and where to get from the equation

$\left\vert w \sum_{n=1}^\infty g(nw) - \int_L g(u) \ du \right\vert \leq tV$,

where $V$ is the total variation of $g(u)$ and $t>0$. Again, I would be grateful for some hints or some literature to dig into - I guess, some knowledge about complex analysis is needed here.

Answer 1

Just reversal of order of summation: $$\ln f(z)=\sum_{n=1}^\infty\ln\frac{1}{1-z^n} =\sum_{n=1}^\infty\sum_{m=1}^\infty\frac{z^{nm}}{m} =\sum_{m=1}^\infty\frac1m\sum_{n=1}^\infty z^{nm} =\sum_{m=1}^\infty\frac1m\frac{z^m}{1-z^m}. $$

Answer 2

\begin{eqnarray*} \log(f(z)) &=&- \sum_{n=1}^{\infty} \log(1-z^n) \\ &=& \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{z^{nm}}{m} \\ \end{eqnarray*} Now invert the order of the plums and sum the geometric series \begin{eqnarray*} = \sum_{m=1}^{\infty}\frac{1}{m} \sum_{n=1}^{\infty} z^{nm} = \sum_{m=1}^{\infty}\frac{1}{m} \frac{z^m}{1-z^m}. \end{eqnarray*}