The following is written in Example 2.3 from Field Theory by JS. Milne:
Let $F$ be a field of characteristic $p \neq 0$ and let $a \in F \setminus F^{p}$. Then $X^p-a$ is irreducible
I fail to see how this holds. I also assume that the authors means it's irreducible in $F[X]$. Also if $\beta$ is a root of this polynomial (in some extension field $\Omega$), then how is $X^p - a = (X - \beta)^p$ in $\Omega$?
Let $a \in F$ such that $a$ does not have a $p^{th}$ root in $F$. We know that, in $\Omega$, $$x^p-a = (x-\sqrt[p]{a})^p.$$
To show that $x^p-a$ is irreducible in $F[x]$, suppose that $x^p-a = f(x)g(x)$ where $f(x),g(x) \in F[x]$. Then $x^p-a = f(x)g(x)$ in $\Omega$ as well; since fields are unique factorization domains and $x-\sqrt[p]{a}$ is irreducible, this implies that $f(x) = (x-\sqrt[p]{a})^i$ and $g(x) = (x-\sqrt[p]{a})^{p-i}$ with $0\leq i \leq p$.
But $$f(x) = (x-\sqrt[p]{a})^i = x^i -i \cdot \sqrt[p]{a}x^{i-1} + \dots \pm \sqrt[p]{a}^i.$$ We assumed that $\sqrt[p]{a} \not \in F$, and that $f(x) \in F[x]$. But look at the coefficient of $x^{i-1}$ in $f(x)$. If $i$ is not divisible by $p$, then $i$ is a unit (F is a field), which would imply that $\sqrt[p]{a} \in F$. Thus it must be that $i$ divides $p$. By our original assumption that $0\leq i \leq p$, this means that either $f(x)$ is a constant polynomial or a scalar multiple of $x^p-a$.
Thus we have shown that $x^p-a$ is irreducible.