Why does $K = \mathbb{F}_p [\theta_n]$ contains subfield of order $p^e$ with solutions to $X^{pe}-X = 0$ over $K$ where $e$ is the exponent of $p\pmod n$?
The setting is as follows. We have $p$ prime, $n$ with $p\not \mid n$, and $e$ the exponent of $p$ modulo $n$ (smallest number $e$ s.t. $ p^e \equiv 1 \pmod n$).
We look at a primitive $n$-th root of unity $\theta$ over $\mathbb{F}_p$ and $K = \mathbb{F}_p [\theta]$. What I don't understand is substep of Donald McCarthy "Sylow's Theorem is a Sharp Partial Converse to Lagrange's Theorem" (1970) Math. Z. 113 S, 383-384.
We have $| K |= p^r$ when $ | K : \mathbb{F}_p | = r$, then $K^\times$ has $p^r-1$ elements. But we know $\theta \in K$ has order $n$ and which implicates $ p^r-1 \equiv 0 \pmod n \Leftrightarrow p^r \equiv 1 \pmod n$ from which follows $ e \mid r$ (since $e$ was the smallest such integer).
Here comes the part I don't understand: $K$ contains a intermediate(subextension) $L$ of order $p^e$ (I guess that part is OK, since we have to subfieldextensions for all dividing orders right?) which consists of all solutions of $X^{pe}-X = 0 $ in $K$. I can't make sense of the "consists of all solutions of $X^{pe}-X$" part, I don't know which part of Field-Theory im forgetting here. But if we have that, then it's easy to see that $\theta$ allready fulfills that condition then $L$ contains $K$, so $L=K$ follows and with that $e=r$.
Thanks for your help!
firstly I think it should be $$X^{p^{e}}-X=0\quad instead\quad of\quad X^{pe}-X=0.$$
Now it is easy cause the solutions of $X^{p^{e}}-X=0$ on $\mathbb{F}_p$ generate a finite field of the same cardinality of $L$ and all elements of $L$ satisfy $X^{p^{e}}-X=0$. Indeed $$X^{p^{e}}-X=X*(X^{p^{e}-1}-1)$$ and every element in $L^{*}$ has order $p^{e}-1$.