$f(x)$ is separable since its derivative is $f'(x) = -1 \ne 0$.

Suppose $\theta$ is a root of $f(x) = x^p - x + a$. Using the Frobenius automorphism, we have:
\begin{align}
f(\theta + 1) &= (\theta + 1)^p - (\theta + 1) + a\\
&= \theta^p + 1^p - \theta - 1 + a\\
&= \theta^p - \theta + a\\
&= f(\theta) = 0
\end{align}

Thus, by induction, if $\theta$ is a root of $f(x)$, then $\theta + j$ is also a root for all $j \in \mathbb F_p$.

By above, if $f(x)$ were to have a root in $\mathbb F_p$, then $0$ would a be a root too, but this contradicts $a \ne 0$. Thus, $f(x)$ has no roots in $\mathbb F_p$. (This can also be shown using Fermat's little theorem.)

Suppose $\theta$ is a root of $f(x)$ in some extension of $\mathbb F_p$. We know that $\theta + j$ is also a root for all $j \in \mathbb F_p$. Since $f(x)$ is of degree $p$, these are all of the roots of $f(x)$.

Clearly, $\mathbb F_p(\theta) = \mathbb F_p(\theta + j)$ for all $j \in \mathbb F_p$. Thus, all $\{\theta + j\}$ have the same degree over $\mathbb F_p$. Since $f(x)$ is separable, it follows that $f(x)$ must be the product of all minimal polynomials of $\{\theta + j\}$. Suppose the minimal polynomials have degree $m$. We have $p = km$ for some $k$. Since $p$ is prime, either $m = 1$; hence $\theta \in \mathbb F_p$, a contradiction. Or $k = 1$; hence $f(x)$ is irreducible because it's the minimal polynomial.