Root of unity over a finite field ($E(n,q)=\{a\in {\overline{\mathbb{F}}}_q:a^n=1\}$)

These roots are taken in an algebraic closure, not merely in ${\mathbb{F}}_q$. If the characteristic of a field is $p$ and $n=p^{a}m$ with $p\nmid m$, then $X^n-1=(X^m-1{)}^{p^a}$; hence there are $m$ distinct roots. They form a cyclic group by the finite multiplicative subgroup of a field is cyclic theorem. In particular they have order $n$ when $\gcd (n,q)=1$.

With $m$ the characteristic-prime-free part of $n$, all roots of unity over a finite field lie in ${\mathbb{F}}_{q^s}$ exactly when $m\mid q^s-1$. This follows from the cyclicity of the multiplicative group of a finite field. For $m=1$ the least positive degree is one. A primitive root of order $m$ has the same field degree, because its Frobenius orbit closes exactly when $m\mid q^s-1$.