$\sum\limits_p\log\left(\frac{1}{1-\frac{1}{p^s}}\right)=\sum\limits_p \sum\limits_{n\geq 1}\frac{1}{np^{ns}}$

Question

I am not seeing the second equality (after Euler's product) of this equality of the Riemann $\zeta$-function for $\operatorname{Re}(s)>1$ $$\log(\zeta(s))=\sum\limits_p\log\left(\frac{1}{1-\frac{1}{p^s}}\right)=\sum\limits_p \sum\limits_{n\geq 1}\frac{1}{np^{ns}}$$ where $p$ means summation over all primes. I tried writing the logarithm as Maclaurin series but this gives me a minus sign in front of the inner sum: Let $q:=\frac{1}{1-\frac{1}{p^s}}$ then $$\sum\limits_p\log\left(1-1+ q\right)=\sum\limits_p \left(-\sum\limits_{n\geq 1}\frac{1}{n}(1+q)^n\right)$$ If I use the identity for $\log(1+(q-1))$ I have this alternating $(-1)$ term in front. I also tried to write the $q$ as a geometric series but then I don't see how I can eliminate the logarithm: $$\sum\limits_p\log\left(\frac{1}{1-\frac{1}{p^s}}\right)=\sum\limits_p\log\left(\sum\limits_{n\geq 1}(p^{-s})^n\right)$$

Answer 1

\begin{eqnarray*} \log\left(\frac{1}{1-\frac{1}{p^s}}\right)=- \log \left(1-\frac{1}{p^s} \right). \end{eqnarray*} Now expand using \begin{eqnarray*} - \log (1-u) = \sum_{n=1}^{\infty} \frac{u^n}{n}. \end{eqnarray*}