 Solution

The multiplicative group of $L={\mathbb{F}}_{q^N}$ is cyclic of order $m=q^N-1$. For a primitive element $\zeta$, $L={\mathbb{F}}_q(\zeta )$, so this is a cyclotomic extension. The cyclotomic character isIts image is the cyclic subgroup $⟨q⟩$, of order $N$.