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.
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!)