Let $\omega(n)$ denote the number of prime factors of a positive integer $n$. Prove that \begin{equation}\sum_{n=1}^{\infty}\frac{2^{\omega(n)}}{n^s}=\frac{\zeta^2(s)}{\zeta(2s)}\end{equation}
My attempt: note that $2^{\omega(n)}$ is multiplicative, since $\omega(n)$ is clearly additive. Therefore the Dirichlet series on the LHS of the question admits an Euler product as follows: \begin{align}\sum_{n=1}^{\infty}\frac{2^{\omega(n)}}{n^s}&=\prod_p\left(\sum_{\nu=0}^{\infty}\frac{2^{\omega(p^k)}}{p^{ks}}\right)\\&=\prod_p\left(\sum_{\nu=0}^{\infty}\frac{2^{k}}{p^{ks}}\right)\\&=\prod_p\left(\sum_{\nu=0}^{\infty}\left(\frac{2}{p^s}\right)^k\right)\\&=\prod_p\left(\frac{p^s}{p^s-2}\right)\\&=\prod_p\left(1+\frac{2}{p^s-2}\right).\end{align}
I can't see how to deduce the result though. The solution in the book (Ram Murty Problems in Analytic Number Theory) seems to follow the same outline as my attempt, but in a slightly different way which I don't quite understand. Their solution is that since $2^{\omega(n)}$ is multiplicative, we have \begin{align}\sum_{n=1}^{\infty}\frac{2^{\omega(n)}}{n^s}&=\prod_p\left(1+\frac{2}{p^s}+\frac{2}{p^{2s}}+\cdots\right)\\&=\prod_p\left(1+\frac{2}{p^s}\left(1-\frac{1}{p^s}\right)^{-1}\right)\\&=\prod_p\left(1-\frac{1}{p^s}\right)^{-1}\left(\frac{2}{p^s}+\left(1-\frac{1}{p^s}\right)\right)\\&=\prod_p\left(1-\frac{1}{p^s}\right)^{-1}\left(1+\frac{1}{p^s}\right)\\&=\zeta(s)\prod_p\left(1+\frac{1}{p^s}\right).\end{align}
I don't really understand the strategy behind their proof; what are they doing? Is there a way to go from my working out to the solution?
As commenters have pointed out, "the number of prime factors of $n$" can be ambiguous: does it mean the number of distinct prime factors, or the number of prime factors counted with multiplicity.
It turns out that it's currently standard in analytic number theory to use $\omega(n)$ to denote the number of distinct prime factors of $n$, and to use $\Omega(n)$ to denote the number of prime factors of $n$ counted with multiplicity.
So the obstacle is that the OP actually worked (correctly) with $\Omega(n)$, while the intended question and solution are working with $\omega(n)$.