Let $\mathbb{F}_Q$ be an extension field of $\mathbb{F}_q$. Show that if an element $a\in \mathbb{F}_Q$ is a root of a polynomial $h(x)\in\mathbb{F}_q[x]$, then so are the elements $a^{q^r}$ for $r\geq0$.
First since $h(x)\in\mathbb{F}_q[x]$ and $\mathbb{F}_q$ is a subfield of $\mathbb{F}_Q$ we have $h(x)\in\mathbb{F}_Q[x]$. For $r=0$ it evidently holds. Can you give some ideas for $r>0$?
EDIT Based on Stahl's suggestions I have proceeded to the following:
Let $h(x)=\sum\limits_{i=0}^mb_ix^i$ for $b_i\in\mathbb{F}_q$, $\mathrm{degree}(h(x))=m$.
Now $a^q\in\mathbb{F}_Q$ since $a\in\mathbb{F}_Q$. Then
\begin{eqnarray} h(a^q)&=&\sum\limits_{i=0}^mb_ia^{qi}\\ &=&\sum\limits_{i=0}^mb_i^qa^{qi} \quad\text{since $b_i^q=b_i$ for } b_i\in\mathbb{F}_q\\ &=&\sum\limits_{i=0}^m(b_ia^{i})^q\\ &&\text{by property $(a+b)^{p^n}=a^{p^n}+b^{p^n}$ and the order of the field can be expressed as some}\\ &&\text{power of its characteristic $q=p^n$}\\ &=&\left(\sum\limits_{i=0}^mb_ia^{i}\right)^q\\ &=&(h(a))^q\\ &=&0 \end{eqnarray}
So for $r=1$ it holds (induction basis). Assume it holds for $r=n$ i.e. $h(a^{q^n})=0$ for $n>1$ (induction hypothesis). For $r=n+1$: $$h(a^{q^{n+1}})=\sum\limits_{i=0}^mb_ia^{q^{n+1}i}=\sum\limits_{i=0}^mb_i^q(a^{q^{n}i})^q=\sum\limits_{i=0}^m(b_ia^{q^{n}i})^q=\left(\sum\limits_{i=0}^mb_ia^{q^{n}i}\right)^q=0$$ verifying the induction step.
Hint 1: In characteristic $p,$ $(a + b)^p = a^p + b^p.$ Use Lagrange's theorem on $\Bbb F_q^\times$ for $r = 1$ and finish the proof via induction on $r.$
Hint 2: Raise both sides of the equation $h(a) = 0$ to the $q^r$-th power and use the above hint.