Question Is there a simple combinatoric proof that in $\mathbb{F}_q[x]$ :$\ \ \displaystyle \boxed{m \, N_m = \sum_{d | m} \mu(d) q^{m/d}}$ ?
The only proof I saw uses formal series :
Let $M$ be the set of monic polynomials, $m_d = \#\{ f \in M, deg(f) = d\}$
$P$ the set of prime/irreducible monic polynomials, $N_d= \#\{ f \in P, deg(f) = d\}$
Then by unique factorization $$\sum_{f \in M} t^{deg(f)} =\sum_{d=0}^\infty m_d t^d = \prod_{h \in P} (1+\sum_{k=1}^\infty t^{deg(h^k)}) = \prod_{d=1}^\infty \frac{1}{(1-t^d)^{N_d }}$$
Taking the logarithm and using that $m_d=q^d$ we have $$-\log(1-qt) = \log(\sum_{d=0}^\infty q^{d} t^d) = -\sum_{d=1}^\infty N_d\log(1-t^d)$$
Using the series for $−\log(1−x)$ : $$\sum_{m=1}^\infty \frac{q^m}{m}t^m =\sum_{d=1}^\infty N_d\sum_{k=1}^\infty \frac{t^{dk}}{k}=\sum_{m=1}^\infty t^m \sum_{d | m} N_d\frac{d}{m}$$
Hence $q^m=\sum_{d | m}d \, N_d$ and by Möbius inversion $$m \, N_m = \sum_{d | m} \mu(d) q^{m/d}$$
Thanks to the link given by PeterHumphries :
in $m \, N_m = \sum_{d | m} \mu(d) q^{m/d}$
$m N_m$ is the number of roots of irreducible polynomials of degree $m$
while the RHS is the number of elements $a \in \mathbb{F}_{q^m}$ such that $\mathbb{F}_{q^m} = \mathbb{F}_{q}(a)$, i.e. the number of elements in $\mathbb{F}_{q^m}$ once we removed the subfields : $\mathbb{F}_{q^d}, d |m$
The formula in the RHS (inclusion-exclusion principle) is based on $\# \mathbb{F}_{q^d} = q^d$ and $\mathbb{F}_{q^d} \subset \mathbb{F}_{q^e}$ iff $d | e$