$p$ is a prime number, $\, n\ge1 \, $and$ \, n \nmid p^n$.If $a \in GF(p)$, proof $x^{p^n}-x-a$ is reducible in $GF(p)$

Question

I know that in a field $F$ with character $p$($p$ is a prime number), we have $x^{p^n}-x-a \,(a \in F)$ is reducible \iff $x^{p^n}-x-a$ has root in $F$ Then the question is to prove $x^{p^n}-x-a \, \equiv \, 0 \, $(mod $p$) has solution.