Showing $f\in\mathbb{F}_{p^d}[X]:f'=0\Rightarrow\exists g\in\mathbb{F}_{p^d}[X]:f=g^p$

Question

Let $\mathbb{F}_{p^d}$ denote the final field with $p^d$ elements and $\mathbb{F}_{p^d}[X]$ denote the polynomial ring in $X$ over $\mathbb{F}_{p^d}$. How can we show $$f\in\mathbb{F}_{p^d}[X]:f'=0\Rightarrow\exists g\in\mathbb{F}_{p^d}[X]:f=g^p$$ The case $d=1$ is easy (since $a^p=a$ for all $a\in\mathbb{F}_p$) and also gives us the other direction. But what about $d\ge 2$?

Answer 1

Hint: You need that if $a\in F_{p^d}$ then $(a^{p^{d-1}})^p = a^{p^d}=a$.