What order is the field: $\mathbb{F}_p(\zeta_{p-1}^{1/n})$?
$\mathbb{F}_p^{\times}$ is size $p-1$ and cyclic (with a generator we shall call $\zeta_{p-1}$). Naively, it seems that by taking the $n^{th}$ root of $\zeta_{p-1}$ we get a group of units of size $C_{n(p-1)}$. This, of course, doesn't make sense. What should be true is that $C_{n(p-1)}$ is some subgroup of the group of units of the $\mathbb{F}_{p^r}$ we get.
Question
What is the order of the field $\mathbb{F}_p[x]/(m_{\zeta_{p-1}^{1/n}})$ where $m_{\zeta_{p-1}^{1/n}}$ is the minimal polynomial of $\zeta_{p-1}^{1/n}$ (which, unless I'm mistaken, is the same as the splitting field of $x^n-\zeta_{p-1}$)? Does it depend on whether $p|n$ or not?
Every nonzero element of a finite field is a root of unity. Roots of unity behave very nicely, since the other roots of unity of the same order are just powers of the first.
For instance i is a primitive 4th root of unity, and the other one, -i, is just a power i,-1,-i,1.
The roots of unity form these nice cyclic subgroups of the group of units of a field.
If you want one primitive k'th root of unity, you get all of them for free, so we are just interested in the smallest finite field of characteristic p with a primitive k'th root of unity.
Now you get into trouble if p divides k: a finite field of characteristic p has size p^m, so its group of units has order p^m-1, and so k has to divide p^m-1. Such numbers are coprime to p.
In your question, you'll want to assume n is coprime to p, otherwise you won't get very good roots. For instance, when p=2, the 8th roots of unity are exactly {1}; none of them are primitive.
Groups of units of finite fields are very well behaved: they are all cyclic. So to find a primitive k'th root of unity, you just want to find the smallest m such that k divides p^m-1:
Note that k has to be coprime to p in order for a primitive k'th root of unity to exist. Just divide k by p until it is coprime if you want to adjoin the "closest to primitive" root.