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?

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 $\mathbb{F}_{2}$. 

