$\mathbb{F}_{p^n} \subset \mathbb{F}_{p^m} \iff m\mid n$ I have been able to show that
$$
x^{p^m} - x\mid x^{p^n} - x \iff m\mid n
$$
My notes say that
$$
\mathbb{F}_{p^n} \subset \mathbb{F}_{p^m} \iff m\mid n
$$
follows from the first fact but I don't understand why. Thanks! 
 A: At first we show that $m|n$ iff $x^m -1|x^n -1.$ ( I will call it a lemma)
Proof. Suppose $m|n.$ Then $n = mq$ for some positive integer $q.$ Then observe that
$x^n -1= x^{mq}-1= (x^m)^q -1 = (x^m-1)((x^m)^{q-1}+.........+1).$ This shows $x^m -1|x^n -1.$
Conversely, suppose $x^m -1|x^n -1.$ Let $n=mq +r.$ We'll show that $r$ has to be zero.
We have $x^n -1= x^{mq +1}-1 = (x^{mq +1}-x^r) + (x^r -1) = x^r(x^{mq}-1) +(x^r-1).$ Now, we conclude that $x^r -1 =0.$ This gives $r=0.$ Hence, we're done. 
Now, let's come back to our original problem. 
Suppose $F_{p^m} \subset F_{p^n}.$ Then by the Lagrange theorem, we have $|F_{p^m}^*|$ divides $|F_{p^n}^*|.$ Note that here we're considering only the multiplicative groups of our given finite fields. Recall that $|F_{p^m}^*|= p^m -1$ and $|F_{p^n}^*|= p^n -1$. Therefore, 
$p^m -1|p^n -1,$ and hence $m|n$ by the lemma. So, are done with the easy direction.
Conversely, suppose $m|n$. Therefore, $x^m -1|x^n -1.$ Please note that in an Example of Dummit and Foote (I think page 549 of 3^rd edition) it was shown that $F_{p^n}$ is the splitting field of $x^{p^n} -x$ over $F_p.$
Now, $x^{p^n}-x = x(x^{p^n -1} -1) = (x^{p^m} -x)$ (some other terms). You've mentioned that you did this part. So,this computation shows that  the roots of the polynomial $x^{p^m} -x$ are included in the roots of $(x^{p^n} -x).$ Therefore, the splitting field of of  $x^{p^m} -x$ contained in the splitting field of  $(x^{p^n} -x)$ ie $F_{p^m} \subset F_{p^n}.$ 
This completes the proof.
