Problem
Prove that for $\operatorname{Re}(s)> 0$, $$ \sum_p {1 \over p^s} = \sum_{n=1}^\infty {\mu(n) \over n} \log \zeta(ns), $$ where the sum extends over all primes $p$.
Notes: $\log$ is natural, $\mu$ is the Möbius $\mu$ function, $\zeta$ is the Riemann $\zeta$ function.
Progress
Using the Euler product for $\zeta(s)$ on the right-hand side gives $$ -\sum_{n=1}^\infty {\mu(n)} \sum_p \log (1-p^{-ns}) = -\sum_p \sum_{n=1}^\infty {\mu(n) \over n} \log(1-p^{-ns}). $$ Equating the terms of this sum with the desired left-hand side would be $$ {1 \over p^s} = -\sum_{n=1}^\infty {\mu(n) \over n} \log(1-p^{-ns}) = \sum_{n=1}^\infty {\mu(n) \over n} \sum_{m=1}^\infty {1 \over m} p^{-mns}, $$ where the Taylor series for $\log(1-x^{-1})$ was used for the second equality. This is the only way I can see to introduce sum over the primes on the right-hand side. Based on some numerical tests in Mathematica, I haven't made a mistake thus far, but I'm stuck here.
After that last step, you need to use $$ \sum_{d\setminus n} \mu(d) = [n=1], $$ (e.g., (18) in mathworld article; I also used the Iverson bracket), and rewrite the sum as a sum over integers $mn$ and their divisors $n$: $$ \sum_{n,m\geq1} \frac{\mu(n)}{mn}p^{-mns} = \sum_{k,n\geq1} \frac{p^{-ks}}{k}\mu(n)[n\setminus k] = \sum_{k\geq1}\frac{p^{-ks}}{k}[k=1]. $$ This is a common trick for this sort of thing.