If $X^{p^d}\equiv X\text{ mod }f$ for prime numbers $p,d$, then $f\in\mathbb{F}_p[X]$ with $\deg f=d$ is irreducible over $\mathbb{F}_p$ Let $p$ and $d$ be prime numbers and $f\in\mathbb{F}_p[X]$ with $\deg f=d$ and no roots over $\mathbb{F}_p$. I want to show $$X^{p^d}\equiv X\text{ mod }f\;\;\;\Rightarrow\;\;\;f\text{ is irreducible over }\mathbb{F}_p$$
As Martin Brandenburg mentioned for $q=p^d$ and $n\in\mathbb{N}$, it holds $$(*)\;\;\;X^{q^n}-X=\prod_{p\in\mathbb{F}_q[X]\;:\;p \text{ is monic and irreducible with }\deg p\;\mid\; n}p$$ That means $$X^q-X=\prod_{(X+\alpha)\in\mathbb{F}_q[X]}(X+\alpha)=X^q\cdot o(X^q)$$ What can we conclude from that?

Hint. $X^{p^d}-X$ is the product of irreducible monic polynomials whose degree divides $d$. In fact, these are the minimal polynomials of the elements of $\mathbb{F}_{p^d} = \{a \in \overline{\mathbb{F}_p} : a^{p^d}=a\}$. 
2nd Hint. $d$ is a prime number.
3rd Hint. Assume that $g$ is a monic irreducible factor of $f$. Show that $g$ has degree $d$. Conclude $g=f$.
