 Solution

Let $\alpha$ be a root in a splitting field. The preceding characteristic-$p$ identity gives $X^p-a=(X-\alpha {)}^p$, so there is only one distinct root. If the polynomial were reducible over $F$, choose a monic proper factor $g\in F[X]$ of degree $d$, with $1\le d<p$. In the splitting field it must be $g=(X-\alpha {)}^d$. Its coefficient of $X^{d-1}$ is $-d\alpha$, which belongs to $F$. Since $d$ is nonzero and thus invertible in the field $F$, this forces $\alpha \in F$, contradicting the absence of a root there. Therefore $X^p-a$ is irreducible over $F$.