Prove that if $x^p - a$ is reducible over a field $F,$ then it has a root in $F.$

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.