# Prove that $\sum_{n=1}^\infty \frac{\sigma_a(n)}{n^s}=\zeta(s)\zeta(s-a)$

I would appreciate a hint concerning how to surpass the roadblock I've encountered in my attempt at a proof below. A nicer proof than mine would also help (Edit: The latter part is now done by Gerry Myserson, the prior remains).

Attempt at a proof (below):

As $\sigma_a(x)$ is completely multiplicative, we can take the infinite product of prime series: $$\sum_{n=1}^\infty \frac{\sigma_a(n)}{n^s}= \prod_{\text{p prime}}\sum_{k=0}^\infty \frac{\sigma_a(p^k)}{p^{ks}}$$ $$=\prod_{\text{p prime}}\sum_{k=0}^\infty \frac{\frac{p^{(k+1)a}-1}{p^a-1}}{p^{ks}}$$ $$=\prod_{\text{p prime}}\sum_{k=0}^\infty \frac{1}{p^a-1}\left[\frac{p^{(k+1)a}}{p^{ks}}-\frac{1}{p^{ks}}\right]$$ $$=\prod_{\text{p prime}} \frac{1}{p^a-1}\left[p^a\zeta(s-a)-\zeta(s)\right]$$ $$=\zeta(a)\prod_{\text{p prime}} \left[\zeta(s-a)-p^{-a}\zeta(s)\right]$$ I cannot see how to extract $\zeta(s)\zeta(s-a)$ from this.

Since $\sigma_a=f\ast g$ is the Dirichlet product with $f(n)=1$ and $g(n)=n^a$, and since multiplication of Dirichlet series is given with this product, we obtain $$\sum_{n=1}^{\infty}\sigma_a(n)n^{-s}=\sum_{n=1}^{\infty}n^{-s}\sum_{n=1}^{\infty}n^an^{-s}=\zeta(s)\zeta(s-a).$$
$$\zeta(s)\zeta(s-a)=\sum_j(1/j^s)\sum_k(k^a/k^s)=\sum_n c(n)/n^s$$ where we have to prove $c(n)=\sigma_a(n)$. But every factorization $n=jk$ contributes $k^a$ to the coefficient of $n^{-s}$.