I am wondering if my solution to the following problem has any issues:
First, we check the case when $char\ F = q \neq 0$: We see $x^q - \alpha^q = (x - \alpha)^q$. Assume for contradiction that this is reducible over $F$. Then we have $(x - \alpha)^n \in F[x]$ for some $n < q$. Choose $n$ minimal. Then $q = nm + r$ for some $m \in \mathbb{N}$ and $1 \leq r < n$. Since $(x - \alpha)^n$ and $(x - \alpha)^q$ are both in $F[x]$, we have that $\alpha^n, \alpha^q \in F$. Then $$ \alpha^q = (\alpha^n)^m \alpha^r\\ \longrightarrow \alpha^r = \frac{\alpha^q}{(\alpha^n)^m} \in F $$ which is a contradiction since $r < n$ and we chose $n$ minimal such that $\alpha^n \in F$.
Next, we assume $char\ F \neq q$. Since $\gcd(x^q - \alpha^q, qx^{q-1}) = 1$, we have that $x^q - \alpha^q$ is separable. So it has $q$ roots.
Let $S = \{\alpha \xi^k | k = 0,1,\ldots, q-1\}$ where $\xi$ is a primitive $q^{th}$ root of unity. We have that $S \in \overline{F} \setminus F$, since by assumption $\xi^k \in F$ for each $k$ and $\alpha \notin F$.
Furthermore, the elements in $S$ are different, and we see that each one is a root of $x^q - \alpha^q$. Thus $S$ is the complete set of roots of $x^q - \alpha^q$. These roots are all contained in $F(\alpha)$.
Since $F(\alpha)$ is the splitting field of a separable polynomial in $F$, we have that $F(\alpha)/F$ is Galois. Let $G = Gal(F(\alpha)/F)$) be the Galois group
We claim that the function $\phi: F(\alpha) \to F(\alpha)$ characterized by $(\alpha \mapsto \xi \alpha)$ is a member of the Galois group. It is easy to see that it is injective and surjective. Furthermore, it fixes $F$.
Consider the polynomial $\pi(x) = \prod_{\{\sigma(\alpha): \sigma \in G\}} (x - \sigma(\alpha))$. By definition, $\pi(x)$ is the minimal polynomial of $\alpha$, so we have $\pi(x) | (x^q - \alpha^q)$. However, each of the $q$ roots of $x^q - \alpha^q$ is also a root of $\pi(x)$, since they are each of the form $\alpha \xi^k$, and such elements can be obtained by repeatedly iterating the Galois automorphism $\phi$ on $\alpha$.
Hence $(x^q - \alpha^q) | \pi(x)$. Since the minimal polynomial is irreducible by definition, so is $x^q - \alpha^q$. Then, since $\deg(\pi(x)) = \deg(\alpha) = q$, we have that $[F(\alpha) : F] = q$.
