I know from this link: $\zeta_m(s)=\prod\limits_{p\nmid m} \frac{1}{\left(1-\frac{1}{p^{f(p)s}}\right)^{g(p)}}$ is a Dirichlet series with non-negative coefficients why $\zeta_m(s)$ is a Dirichlet series:
If $A(s)=\sum_n a_nn^{-s}, B(s) =\sum_n b_nn^{-s}$ we have
$$A(s)B(s)=\sum_n\left(\sum_{d|n}a_d b_{n/d}\right)n^{-s}$$
That is, the product of 2 Dirichlet series of the forms $A$ and $B$ is a Dirichlet series. Since $\zeta_m(s)$ is exactly such a finite product it is a Dirichlet series.
However, I do not understand how $\zeta_m(s)$ would have positive coefficients. I didn't quite get the expiation provided in the link.
Can someone care to carefully explain why $\zeta_m(s)$ would have positive coefficients? I just can't seem to see how the form of $\zeta_m(s)$ in the picture above somehow allows us to conclude that it Dirichlet Series has positive coefficients. Thanks
Maybe it's easier to see if you write it like $$\zeta_m(s) = \prod\limits_{p \nmid m}(1+p^{-f(p)s} + p^{-2 f(p)s} + \cdots)^{g(p)}$$ Let me do the special case when all the $g(p)$ are equal to $1$. If $n$ is an integer relatively prime to $m$, write $n = \prod\limits_p p^{e_p}$ and set $$h(n) = \sum\limits_p e_p f(p)$$ Then
$$\zeta_m(s) = \sum\limits_{n \in \mathbb N, (n,m) = 1} \frac{1}{n^{h(n)s}} $$
When the $g(p)$ are positive terms, you must expand $(1+p^{-f(p)s} + p^{-2f(p)s} + \cdots)^{g(p)}$ first before continuing.