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

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).
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.