Prove $\sum_{n=1}^{\infty}\frac{\nu(n)}{n^s}=\zeta(s)\sum_p\frac{1}{p^s}$. How to prove   $\sum_{n=1}^{\infty}\frac{\nu(n)}{n^s}=\zeta(s)\sum_p\frac{1}{p^s}$.
where $\nu(1)=0$ and $\nu(n)=k$ if $n=p_1^{a_1}p_2^{a_2}p_3^{a_3}\cdots p_k^{a_k}$
for $\sigma>1$ where $\sigma$ is real part of s.
This exercise is from Tom M. Apostol, Introduction to Analytic Number Theory book chapter 11, problem number 6.
There is a theorem in the book which says if $\sum f(n)n^{-s}$ converges absolutely for $\sigma>\sigma_a$.If f is multiplicative then we have
$\sum_{n=1}^{\infty} f(n)n^{-s}=\Pi_p({1+\frac{f(p)}{p^s}+\frac{f(p^2)}{p^{2s}}+ \cdot\cdot\cdot)}$
but here $\nu(n)$ is not multiplicative rather additive.So I can't use this theorem.
Any hint will be appreciated.

Hint: the multiplication of Dirichlet generating functions corresponds to the generating function of the Dirichlet convolution
