Let $\alpha \in \overline{\mathbb{F}_{p}}$ be a root of $x^{p} - x + a$. To show $x^{p} - x + a$ is irreducible, it suffices to show that $m_{\alpha, \mathbb{F}_{p}}(x) = x^{p} - x + a$. Moreover, since $m_{\alpha, \mathbb{F}_{p}}(x) \mid (x^{p} - x + a)$, it suffices to show that $\deg m_{\alpha, \mathbb{F}_{p}}(x) \geq p$. We also know that $[\mathbb{F}_{p}(\alpha) : \mathbb{F}_{p}] = \deg m_{\alpha, \mathbb{F}_{p}}(x)$, so we only need to prove that $[\mathbb{F}_{p}(\alpha) : \mathbb{F}_{p}] \geq p$. This can be done by showing that $$ p \leq \lvert \mathrm{Aut}(\mathbb{F}_{p}(\alpha) / \mathbb{F}_{p}) \rvert \leq \#\{ \mathbb{F}_{p}\text{-embeddings }\varphi : \mathbb{F}_{p}(\alpha) \to \overline{\mathbb{F}_{p}} \} \leq [\mathbb{F}_{p}(\alpha) : \mathbb{F}_{p}]. $$ $\operatorname{Fr}$ is a good instance of $\mathrm{Aut}(\mathbb{F}_{p}(\alpha) / \mathbb{F}_{p})$. If $\operatorname{id}_{\mathbb{F}_{p}(\alpha)}, \operatorname{Fr}, \dots, \operatorname{Fr}^{p-1}$ are different, then of course $\lvert \mathrm{Aut}(\mathbb{F}_{p}(\alpha) / \mathbb{F}_{p}) \rvert \geq p$ for that $\mathrm{Aut}(\mathbb{F}_{p}(\alpha) / \mathbb{F}_{p})$ is a group. Let us check it on $\alpha$. First, since $\alpha$ is a root of $x^{p} - x + a$, we have $$ \operatorname{Fr}(\alpha) = \alpha^{p} = \alpha - a. $$ Then $$ \operatorname{Fr}^{2}(\alpha) = \operatorname{Fr}(\alpha - a) = \operatorname{Fr}(\alpha) - \operatorname{Fr}(a) = \alpha - a - a^{p} = \alpha - 2a. $$ Using induction on $n$, we can prove that $\operatorname{Fr}^{n}(\alpha) = \alpha - na$. Therefore, $\operatorname{id}_{\mathbb{F}_{p}(\alpha)}(\alpha), \operatorname{Fr}(\alpha), \dots, \operatorname{Fr}^{p-1}(\alpha)$ are differentiable, so are those automorphisms.