Let $F = \mathbb{F}_{p^n}$ be a field of $p^n$ elements. For $1\leq k \leq n$ set $L_k = \{a \in F: a^{p^k} = a\}$. I want to show $\{L_k: 1 \leq k \leq n\}$ is the set of all subfields of $F$, and for $n$ greater than $2$, $L_i = L_j$ for some $1 \leq i < j \leq n$.
I'm not sure how to approach this. Some hint would be greatly appreciated.
Notation: $K^\times = K \setminus \{0\}$ for a field $K$.
First part: Let $K$ be a subfield of $F$. Show that $|K| = p^k$ for some $k$ with $1 \leqslant k \leqslant n$.
Note that $K^\times$ is group with order $|K| - 1$. Thus, every element $a \in K^\times$ satisfies $a^{p^k - 1} = 1$. Show that $K = L_k$.
(Recall that the polynomial $x^{p^k} - x$ cannot have more than $p^k$ roots.)
Second part: Show that if $j \nmid n$, then $F$ cannot have a subfield of cardinality $p^j$.
Picking any such nondivisor $j$ shows that $L_j$ cannot have $p^j$ elements. So, $L_j$ has strictly fewer elements. (Why?)
By the previous part, conclude that $L_j = L_i$, where $i$ satisfies $p^i = |L_j|$.