In context of a exercise about expressing the dirichlet series $$\sum\limits_{n=1}^\infty \frac{\nu(n)}{n^s}$$ in term of the zeta function, where $\nu(n)$ denotes the amount of different prime divisors of $n$ (e.g. $\nu(3^5\cdot 7)=2$), if found the upper identity $$\sum\limits_{n=1}^\infty \frac{\nu(n)}{n^s} = \sum\limits_{m=1}^\infty \frac{1}{m^s}\sum\limits_p \frac{1}{p^s}$$ that was considered being trivial. However, I spend the last half an hour starring at this and I have no idea why this should be true and even less why it is considered obvious.
I'd appreciate any hint on this, not necessarily a full explanation.
We can write $$\sum_{n} \frac{\nu(n)}{n^s} = \sum_{n} \sum_{p | n}\frac{1}{n^s}.$$ Swapping order of summation $$\sum_{n} \sum_{p | n}\frac{1}{n^s} = \sum_{p} \sum_{p | n} \frac{1}{n^s}.$$ Setting $m = n/p$, we have $$ \sum_{p | n} \frac{1}{n^s} = \sum_{m}\frac{1}{p^sm^s}.$$ So we conclude that $$\sum_{n} \frac{\nu(n)}{n^s} = \sum_{p}\sum_{m}\frac{1}{p^sm^s}= \sum_{m}\frac{1}{m^s}\sum_{p}\frac{1}{p^s}.$$