For a finitely generated commutative ring R define its zeta function to be (where product is taken over all maximal ideals of $R$): $$\zeta_R(s)=\prod_{m\subset R}\frac{1}{1-|R/m|^{-s}}$$ Since, $R$ is finitely generated, we know that $R/m$ is a finite field.
Prove that, $$\zeta_{\mathbb{F}_p[x]}(s)=\frac{1}{1-p^{1-s}}$$
My attempt till now: We know that the only maximal ideals of $\mathbb{F}_p[x]$ are those generated by monic irreducible polynomials. Therefore, over every irreducible and monic polynomial $f \in \mathbb{F}_p[x]$ $$\zeta_{\mathbb{F}_p[x]}(s)=\prod_{f}\frac{1}{1-|\mathbb{F}_p[x]/(f)|^{-s}}=\prod_{f}\frac{1}{1-p^{-s \cdot (\deg{f})}}$$ How do I go from here to the answer, I think I may have to apply $\log$ and then use another result (which is here, as another question put up by me). But a bit lost, any help?
By unique factorization in the PID $\mathbb{F}_p[x]$ $$\prod_{f \text{ monic irreducible}} \frac{1}{1-t^{\deg(f)}} =\prod_f (1+\sum_{k \ge 1} t^{\deg(f^k)})= \sum_{m \text{ monic}} t^{\deg(m)} = \sum_{d=0}^\infty p^d t^d= \frac{1}{1-p t}$$
Taking the logarithm you'll obtain $$\# \{ f \text{ monic irreducible}, \deg(f) = n\} = \frac{1}{n}\sum_{d | n} \mu(d) p^{n/d}$$ Where $\mu$ is the Möbius function