This comes from Lang's algebra, chapter $5$ exercise $23$ $(b)$. Part $(a)$ asks to find a rational function to represent $$Z(t) := (1-t)^{-1} \prod_p (1-t^{\deg p} )^{-1}$$ Where $p$ ranges through all monic irreducible polynomials of a finite field of cardinality $q$ (which I will denote $F_q$). I can solve this - you find $$Z(t) = \frac{1}{1-qt}$$ $\textbf{Main question:}$ Part $(b)$ asks the following: Let $\pi_q (m)$ denote the number of monic irreducibles with degree $\leq m$. Prove $$\pi_q (m) \sim \frac{q}{q-1} \cdot \frac{q^m}{m}$$ I assume that it is worthwhile to use the result of part $(a)$; use the notation $\psi (d)$ to denote the number of monic irreducibles of degree $d$ so that $\pi_q (m) = \sum_{d=1}^m \psi(d)$. If I take the natural log of both sides I see $$\log (1-qt) = \sum_{d=1}^\infty \psi(d) \log(1-t^d)$$ I did this because I get pretty close to having the sum of my $\psi (d)$ terms occur, but if I continue doing standard power series manipulations I essentially do the solution for part $(a)$ backwards and get nowhere.
Another attempt is to "brute force" it. I can look at $\sum_{d=1}^m \psi (d)$, and using the standard formula for $d \psi (d) = \sum_{d \ | \ n} \mu (d) q^{n/d}$, I can manipulate my sum to get it into the form $$\pi_q (m) = \sum_{d=1}^m \frac{\mu (d)}{d} \sum_{k=1}^{[m/d]} \frac{q^k}{k}$$ where $[\cdot]$ denotes floor function. I confess that I am not terribly familiar with asymptotic properties of the Mobius function, so that this form is not particularly enlightening to me, either. Is there any other way to see this relation more clearly? Any help is appreciated.