TITLE: Taking logarithm of a generating function identity.
QUESTION [5 upvotes]: A week or so ago I asked why
$$
\prod_{n\geq 1}\frac{1}{(1-x^n)^{m_n(q)}}=\frac{1}{1-qx}
$$
where $m_n(q)$ the number of irreducible monic polynomials with degree $n$ over the finite field of order $q$.
Why does taking logarithms then imply 
$$
\sum_{n\mid r}nm_n(q)=q^r
$$
for any $r$? I know taking logarithms will change it to an additive identity, but don't see how this particular equality falls out. Thank you.

REPLY [4 votes]: If we take the logarithm and use the series expansion $-\log(1-t) = \sum_{n\geq 1} \frac{x^n}{n}$, we see that
$$\sum_{n \geq 1}m_n \sum_{k \geq 1}\frac{x^{nk}}{k} = \sum_{r \geq 1}\frac{q^rx^r}{r}.$$
Now compare the coefficient of $x^r$ on both sides and it follows immediately that
$$\sum_{n \mid r}nm_n = q^r.$$
Note: the product formula that you have written above is essentially the Euler factorization of the zeta function of the affine line over $\mathbb{F}_q$.
Note 2: By using the Möbius inversion formula, it follows that the number of irreducible polynomials of degree $n$ over $\mathbb{F}_q$ is
$$m_n = \frac{1}{n}\sum_{d\mid n} \mu(d)q^{n/d}.$$
(In particular, this is never $0$, because the $d=1$ term dominates the rest of the sum. Also, it's not even obvious a priori that the RHS should be an integer!)