Minimal polynomial of a power of an element over $\mathbb F_{2}$ Let $\beta \in \mathbb F_{16}$ whose minimal polynomial over $\mathbb F_{2}$ is $x^4+x+1$.
What is the minimal polynomial of $\beta^7$ over $\mathbb F_{2}$?
I know from this answer that $\beta$ generates $\mathbb F_{16}$ but I don't know if that helps at all. Can this be done without first finding out what element $\beta$ is?

I have an other solution (but I'm not sure how to generalize it) :
Let $g \in \mathbb{F}_2[x]$ be the minimal polynomial of $\beta^7$. 
$\mathbb{F}_{16}$ is of characteristic $2$, thus $x \mapsto x^2$ is an automorphism. Therefore $g(\beta^7) =0\implies g((\beta^7)^{ 2^k}) = 0$, ie. the roots of $g$ are : $$\beta^7,(\beta^7)^2= \color{red}{\beta^{-1}}, (\beta^7)^{2^2}=\beta^{-2},(\beta^7)^{2^3}=\beta^{-4}$$
$\beta^{-1}$ generate the same field as $\beta$ so their minimal polynomial are of same degree : $4$
Finally $\beta$ is a root of $f(x) = 1+x+x^4$ so that
$$0=\beta^{-1} f(\beta) = \beta^{3}+1+\beta^{-1}$$
ie. $\beta^{-1}$ is a root of $x+1+x^{-3}$ and hence of $g(x) =x^3(x+1+x^{-3})= x^4+x^3+1$ which has to be its minimal polynomial and the minimal polynomial of $\beta^7$, by the preceding.
