Prove that if $x^p - a$ is reducible over a field $F,$ then it has a root in $F.$ This question was asked on my abstract algebra quiz, and I was unable to solve it.

Consider a field $F$ with an element $a \in F$ and a prime number $p.$ Suppose that the polynomial $x^{p} - a$ is reducible in $F[x].$ Prove that this polynomial has a root in $F.$

Attempt: Reducibility of the polynomial implies that $x^{p}-a =q(x)r(x)$ with $q(x)$ and $r(x)$ both nonzero non-units. Now, I assumed that neither $q(x)$ nor $r(x)$ has a root in $F,$ so $\frac{1}{q(x)}$ and $\frac{1}{r(x)}$ are well-defined for all $x$ in $F.$ But I need to prove that $\frac{1}{q(x)}$ and $\frac{1}{r(x)}$ belong to $F[x].$ They can be defined using the Newton Binomial Theorem, but there is no surety that they would necessarily belong to $F[x].$
So, I think there is a problem in my approach. Can anyone please tell me how to approach the problem? Thanks.
 A: I think the right statement should be

Let $p$ be a prime and $F$ be a field with characteristic $q \neq p$. Then for every $a \in F^{\times}$ the polynomial $f(x)=x^p - a$ either has a root or is irreducible.

At first, $f'(x)=px^{p-1} \neq 0$ since $p \neq q$. Therefore, $f$ is separable. Consider the splitting field of $F$, say $E$. Let $\alpha \in E$ be a root of $f$,  $\alpha^p = a$ and $\omega$ be a primitive element of order $p$ in $E$ then all root of $f$ are $S=\left \{\alpha,\alpha\omega,...,\alpha \omega^{p-1} \right \}$. Suppose $f$ is reducible in $F[x]$ then there exists two polynomial $g,h \in F[x]$ such that
$$x^p - a = g(x)h(x).$$
Consider the same representation, but in $E[x]$, let $0<k=\mathrm{deg}(g)<p$ then
$$g(x) = \prod_{s \in S}(x - s).$$
In particular, $g(0)=\pm \alpha^k \omega^n$, we may assume $g(0)=\alpha^k\omega^n$ so $(g(0))^p = \alpha^{kp}\omega^{np}=a^k$. Since $k<p$ there exists $s,t$ such that $ks + pt = 1$ and hence
$$a = a^{ks+pt}= a^{ks}a^{pt}=(g(0))^{ps}a^{pt}=(a^tg(0)^s)^{p}.$$
But recall that $g(0),a \in F \Rightarrow a^tg(0)^s \in F$ so consequently $f(x)$ has $a^tg(0)^s \in F$ as a root.
A: Let $K$ denote $F(\zeta)$, where $\zeta$ is the $p$th root of unity $e^{2\pi i/p}$. Since $f=x^{p}-a$ is reducible over $F$, it's also reducible over $K$. So $f$ splits completely over $K$.
{$s, \zeta s..., \zeta^{p}s$} are the $p$ roots of $f$. Suppose none of them is in $F$. We can choose an element $\sigma$ of $G=G(K/F)$ different from the identity. Note that $K = F(\zeta)$ is also generated by $s$ and $\zeta s$, so $\sigma$ cannot fix both $s$ and $\zeta s$. WLOG, we assume $\sigma(s) = \zeta^{k}s$, $\sigma^{n}(s)=\zeta^{nk}(s)$. Because $k<p$ and p is a prime, $nk$ can run through all residues modulo $p$. This shows that $G$ operates transitively on the $p$ roots of $f$. However $|G|\leq p-1$, which leads to a contradiction. Therefore some of {$s, \zeta s..., \zeta^{p}s$} must be in $F$.
