I am currently working on a problem that requires me to get find a simpler expresion for: $$\sum_{f \in \mathcal{S}_h} \frac{1}{|f|^s} $$
Where $\mathcal{S}_h$ is the set of $h$-full polynomials (i.e. $f \in \mathcal{S}_h$ if the multiplicity of each prime factor of $f$ is $h$ or greater) over a finite field of cardinality $q$ and where $|f|:= q^{\deg(f)}$.
Using Euler's product one can see that: \begin{equation*} \begin{split} \sum_{f \in \mathcal{S}_h} \frac{1}{|f|^s} &= \prod_{P \ \text{prime}} \left(1+\frac{1}{|P|^{hs}} + \frac{1}{|P|^{(h+1)s}}+\dots\right)\\ &=\prod_{P \ \text{prime}} \left(1+\frac{1}{|P|^{hs}\left(1-\frac{1}{|P|^s}\right)} \right)\\ &= \prod_{P \ \text{prime}} \left(\frac{1-\frac{1}{|P|^s} + \frac{1}{|P|^{hs}}}{1-\frac{1}{|P|^s}}\right) = \zeta_q(s) \ \prod_{P \ \text{prime}} \left(1-\frac{1}{|P|^s} + \frac{1}{|P|^{hs}} \right) \end{split} \end{equation*}
I've thought about how to simplify that last product but I haven't got closer to a cleaner expression. I was expecting to find one similar to that of $h$-free polynomials which is $\frac{\zeta_q(s)}{\zeta_q(hs)}$ but I haven't been able. Any ideas of how to continue? Thanks in advance!