Dirichlet series for $\zeta(s) \zeta(2s)\zeta(3s) \zeta(6s)^{-1}$ Let $\kappa(n)$ be defined, for $n=p_1^{a_1}\dots p_k^{a_k}$, by $\kappa(n)=a_1\cdot a_2\dots \cdot a_k$. Prove that
$$
\sum_{n=1}^{\infty}\frac{\kappa(n)}{n^s}=\frac{\zeta(s)\cdot \zeta(2s) \cdot \zeta(3s)}{\zeta(6s)}.
$$
I would like to prove this statement, however I'm not sure what to do with the $\zeta(n)$ function. Are there any useful properties I can use?
 A: @vitamind's approach notes$$(\kappa\ast1)(p^k)=\sum_{j=0}^kj=\tfrac12k(k+1)$$(except of course $(\kappa\ast1)(p^0)=1$) so, since $\sum_{k\ge0}\tfrac12k(k+1)z^k=\tfrac{z}{(1-z)^3}$,$$\begin{align}\sum_{n\ge1}(\kappa\ast1)(n)n^{-s}&=\prod_{p\in\Bbb P}\left(1+\frac{p^{-s}}{(1-p^{-s})^3}\right)\\\implies\sum_n\kappa(n)n^{-s}&=\prod_p\frac{1-p^{-s}+p^{-2s}}{(1-p^{-s})^2}.\end{align}$$Now just verify$$\frac{1-p^{-s}+p^{-2s}}{(1-p^{-s})^2}=\frac{1-p^{-6s}}{(1-p^{-s})(1-p^{-2s})(1-p^{-3s})}.$$Another approach lacking Dirichlet convolution uses$$1+\sum_{k\ge1}kz^k=\frac{1-z+z^2}{(1-z)^2}$$with $z:=p^{-s}$, after which the calculation is the same as above. @pisco's deleted answer used this.
A: Another slightly different approach is to use the fact that the function $\kappa$ is multiplicative and so we have the Euler Product for the Dirichlet series:
$$\sum_{n=1}^\infty \frac{\kappa(n)}{n^s} = \prod_{p} \sum_{n=0}^\infty \frac{\kappa(p^n)}{p^{ns}} = \prod_p \left(1 + \sum_{n=1}^\infty np^{-ns}\right) = \prod_p \left(1 + \frac{p^{-s}}{(1-p^{-s})^2}\right) = \prod_p \frac{1 - p^{-s} + p^{-2s}}{(1-p^{-s})^2}$$
Now using that $\zeta(s) = \prod_p \frac1{1-p^{-s}}$ you get:
$$\frac{\zeta(s)\zeta(2s)\zeta(3s)}{\zeta(6s)} = \prod_{p} \frac{1-p^{-6s}}{(1-p^{-s})(1-p^{-2s})(1-p^{-3s})} = \prod_p \frac{1 - p^{-s} + p^{-2s}}{(1-p^{-s})^2}$$
As the two Euler products agree you have the wanted identity.
REMARK: This is Exercise $9$ of Chapter $11$ in the well-known Apostol's book "Introduction to Analytic Number Theory" $(1976)$. You can find the proof for the Euler product formula for a Dirichelt series involving a multiplicative function in that chapter.
