 Solution

First make the irreducible polynomial monic; for a nonmonic polynomial restore its leading coefficient in front of the product below. If $\alpha$ is one root, $K={\mathbb{F}}_q(\alpha )$ has degree $n$ and cardinality $q^n$. Thus every element satisfies $x^{q^n}=x$. The Frobenius automorphism $x\mapsto x^q$ fixes ${\mathbb{F}}_q$ and permutes the roots. If the orbit of $\alpha$ had size $d<n$, its orbit product would be a degree-$d$ polynomial fixed by Frobenius, hence with coefficients in ${\mathbb{F}}_q$, contradicting irreducibility. ThereforeThe roots are distinct, since $X^{q^n}-X$ has derivative $-1$. They all lie in $K$ and include its generator $\alpha$, so the splitting field is ${\mathbb{F}}_{q^n}$. Frobenius has order $n$ there; a degree-$n$ extension has at most $n$ automorphisms, so