Prove that every monic polynomial $f$ irreducible in $\mathbb {F}_q[x]$ has $\deg(f)$ roots in a finite extension $ E $ such that $\deg(f) \mid s$ and $[E: \mathbb{F}_q] = s$.
The book I'm reading uses this fact without giving a proof; why is this true?
Since finite fields are perfect, the polynomial $f$ is separable, so it has $\deg(f)$ distinct roots in its splitting field. Actually, in this case the splitting field is $\mathbb{F}_q(a)$, where $a$ is a root of $f$, because of uniqueness of finite fields of a given cardinality.
We also know that $[\mathbb{F}_q(a):\mathbb{F}_q]=\deg(f)$.