I was reading a proof about the uniqueness of each subfield of any given order and came across a line in the proof I couldn't understand.
Let $F_q$ be the finite field with $q = p^n$ elements. Then every subfield of $F_q$ has order $p^m$, where m is a positive divisor of n. Conversely, if m is a positive divisor of n, then there is exactly one subfield of $F_q$ with $p^m$ elements
Proof
Clearly, a subfield K of F must have order $p^m$ for some positive integer $m ≤ n.$ By Lemma 6.1, q = $p^n$ must be a power of p m, and so m must divide n.
Conversely, if m is a positive divisor of n, then $p^{ m −1}$ divides $p ^{n −1}$, and so $x^{p^m−1} −1$ divides $x^{p^n−1} −1$ in $F_p[x]$. So, every root of $x^{p^m}−x$ is a root of $x^{p^n} −x$, and hence belongs to $F_q$. It follows that $F_q$ must contain a splitting field of $x^ {p^m} − x$ over $F_p$ as a subfield, and (from proof of Theorem 6.5) such a splitting field has order $p^ m$. If there were two distinct subfields of order $p^ m$ in $F_q$, they would together contain more than $p^m$ roots of $x^{p^m} − x$ in $F_q$, a contradiction.
The difficulty I'm having with the proof lies in this sentence:
if m is a positive divisor of n, then $p^{ m −1}$ divides $p ^{n −1}$, and so $x^{p^m−1} −1$ divides $x^{p^n−1} −1$ in $F_p[x]$.
I'm not sure how $p^{m-1}|p^{n-1}$ and how $x^{p^m−1} −1|x^{p^n−1} −1$ follows from this.
Any help would be appreciated.
You've copied the text down incorrectly: it says $p^m-1$ divides $p^n-1$, not $p^{m-1}$ divides $p^{n-1}$ (even though that is also a true statement).
There is a general fact, which is easy to prove: if $a\mid b$, then $x^a-1$ divides $x^b-1$, since $$x^b-1=(x^a)^k-1^k=(x^a-1)((x^a)^{k-1}+(x^a)^{k-2}+\cdots+(x^a)+1)$$ So since $m\mid n$, we can apply this fact to get that $p^m-1$ divides $p^n-1$, and then applying that same fact again, we get $x^{p^m-1}-1$ divides $x^{p^n-1}-1$.