Also, we have by the definition of Λ,
$$\sum_{n\geq 1} \Lambda(n) n^{−s} = \sum_p(\log p) \sum_{n \geq 1}p^{−ns}$$
(From https://proofwiki.org/wiki/Logarithmic_Derivative_of_Riemann_Zeta_Function)
This step is apparently so trivial that it needn't be included in the proof, but I cannot work out how it works. I know that it's a relatively straightforward step, but I have no idea...
Because $\Lambda(n)$ is zero unless $n$ is a prime power $p^m$, we can only worry about prime powers, and sum over these instead. So (assuming the sum converges at all), we have $$\sum_{n \geq 1} \Lambda(n) n^{-s} = \sum_{p \mbox{ prime}} \sum_{m \geq 1} \Lambda(p^m) (p^m)^{-s} = \sum_{p \mbox{ prime}} \sum_{m \geq 1} (\log p) p^{-ms} = \sum_{p \mbox{ prime}}(\log p) \sum_{m \geq 1} p^{-ms}. $$ Now make the (unfortunate) substitution of the dummy variable $n$ for $m$ on the right, and you get the claimed formula.