Suppose $p$ is prime and $n|p-1$ (I suppose we could take $n$ to be another prime such that it divides $p-1$. Then I am trying to show $x^n-1$ splits into a product of linear factors over $\mathbb{F}_p$. I feel like it should have something to do with $\text{Aut}(C_p)\cong C_{p-1}$ In particular, if $a$ generates $\text{Aut}(C_p)$ then it has order $p-1$. So we take $b=a^{(p-1)/n}$ which clearly has order $n$. I feel that $x^n-1=(x-1)(x-b)(x-b^2)\cdots(x-b^{n-1})$ but unsure if this is the case, or why it is if so. Certainly, if we plug in some numbers then this is true. For example:
$p=3$ and $n=2$ and clearly $(x^2-1)=(x-1)(x-2)$
$p=7$ and $n=3$ and $(x^3-1)=(x-1)(x-2)(x-4)$
$p=11$ and $n=5$ and $(x^5-1)=(x-1)(x-3)(x-4)(x-5)(x-9)$.
I suppose I could argue $x^n-1$ splits completely because it has $n$ distinct zeroes in $\mathbb{F}_p$. Any help on why this is happening, though?
On the finite field $\mathbb{F}_p$, the polynomial $X^{p-1}-1$ splits, as any nonzero element is a root.
If $n\mid p-1$, then $X^n-1 \mid X^{p-1}-1$ (this is a classical result, actually the divisibility relation holds in $\mathbb{Z}[X]$), and so $X^n-1$ divides a split polynomial : it must be split.
Another way to see it, as a comment points out, is that $\mathbb{F}_p^\times$ is cyclic, and it's known that if $G$ is a cyclic group of order $m$, then for any $n\mid m$ there are precisely $n$ elements whose order divides $n$, that is, who satisfy $x^n = 1$ : since there are $n$ roots to $X^n-1$, this polynomial splits.
(As you probably noticed, this can be extended to any finite field, changing $p$ into $q$ wherz $q= p^k$ is any prime power)