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?
Daniel Schepler's suggestion in the comments is a good way to approach this problem. Let me flesh out this method, as well as a few computational tricks/sanity checks.
First, as you note, $\beta$ is a generator for $\mathbb{F}_{16}^{\times}$, which is a cyclic group of order $15$. Since $\gcd(7, 15) = 1$, it follows that $\beta^{7}$ also generates $\mathbb{F}_{16}^{\times}$, and so is a generator for $\mathbb{F}_{16}$ as a field extension of $\mathbb{F}_{2}$. Thus, we expect the minimal polynomial of $\beta^{7}$ over $\mathbb{F}_{2}$ to have degree $4$.
We need to compute $\beta^{7}, \beta^{14}, \beta^{21}, \beta^{28}$ in terms of the $\mathbb{F}_{2}$-basis $1, \beta, \beta^{2}, \beta^{3}$ by using the relation $\beta^{4} + \beta + 1 = 0$, i.e. $\beta^{4} = \beta+1$. This is easy enough for $\beta^{7}$: $$\beta^{7} = \beta^{4}(\beta^{3}) = (\beta+1)(\beta^{3}) = \beta^{4}+\beta^{3} = \beta^{3} + \beta+1$$ Now, it is not too hard to do this for $\beta^{14}$, etc., but one thing which makes computations quicker is to use the fact that $\mathbb{F}_{16}$ has characteristic $2$, so $(a+b)^{2} = a^{2}+b^{2}$ for any $a, b \in \mathbb{F}_{16}$. Thus, we note $$\beta^{14} = (\beta^{7})^{2} = (\beta^{3}+\beta+1)^{2} = \beta^{6}+\beta^{2}+1 = \beta^{2}(\beta+1)+\beta^{2}+1 = \beta^{3}+1$$ and $$\beta^{28} = (\beta^{14})^{2} = (\beta^{3}+1)^{2} = \beta^{6}+1 = \beta^{2}(\beta+1)+1 = \beta^{3}+\beta^{2}+1$$ From here, it is not too hard to compute $\beta^{21}$ by computing $\beta^{7}\beta^{14}$. I leave this (and the final computation of the minimal polynomial of $\beta^{7}$) to you. You can check these computations here (spoilers):
$\beta^{21} = \beta^{3}+\beta^{2}$. Thus, $\beta$ is a root of $X^{4}+X^{3}+1 \in \mathbb{F}_{2}[X]$. As noted above, the minimal polynomial of $\beta^{7}$ over $\mathbb{F}_{2}$ must have degree $4$, so this must be the minimal polynomial of $\beta$ over $\mathbb{F}_{2}$; however, it is also easy to check directly that this polynomial is irreducible over $\mathbb{F}_{2}$.
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.
A systematic way is to consider the map $x \mapsto \beta^7 x$, write its matrix with respect to the basis $1, \beta, \beta^2, \beta^3$, and then find the minimal polynomial of this matrix. Note that $\beta^{7} = \beta^{3} + \beta+1$.
