Given is the polynomial $f(X)=X^{16}-X \in \mathbb F_2 [X]$.
It factorizes into $f(X)=X(X+1)q(X)$, where $q(X)=X^{14}+X^{13}+\dots +X+1$.
Now, let $\alpha$ be a primitive root of $\mathbb F_{16}$ and define $\beta := \alpha ^3$.
Now the first question is, what are the orders of $\alpha$ and $\beta$ in $\mathbb F_{16}^*$.
$\mathbb F_{16}^*$ is the group of units. Since $\alpha$ is a primitive root, it has to generate $\mathbb F_{16}^*$ and so the order of $\alpha$ is $\varphi(16)=\varphi(2^4)=(2-1)2^{4-1}=8$. Is this right? And what does this give to me for the order of $\beta$?
The second question is: What is the lowest degree of a nontrivial polynomial in $\mathbb F_2 [X]$, which has the root $\beta$. Here I don't know, hoe to go on. I thought about, that if a polynomial has $\beta$ as a root, then it has also the roots $\beta^2,\beta^3,\dots$, but I don't know wheteher this is helpful.
A primitive root of a finite field is a generator of its multiplicative group.
$\alpha$ has order $15$ and so $\beta=\alpha^3$ has order $5$. This is true in any group.
Now, $\beta$ is neither in $\mathbb F_2$ nor in $\mathbb F_4$ because it is not a root of $X^3-1$. Therefore, the smallest field that contains $\beta$ is $\mathbb F_{16}$ and so its degree over $\mathbb F_2$ is $4$.
Indeed, since $x^5-1$ factors as $(x + 1) (x^4 + x^3 + x^2 + x + 1) \bmod 2$, the minimal polynomial of $\beta$ is $x^4 + x^3 + x^2 + x + 1$.