TITLE: $F_p(a)$ contains all the roots of $f$
QUESTION [1 upvotes]: Let $p$ be a prime, $n\in \mathbb{N}$ and $f=x^{p^n}-x-1\in \mathbb{F}_p[x]$ irreducible . 
Let $a\in \overline{\mathbb{F}}_p$(=algebraic closure of $\mathbb{F}_p$) is a root of $f$. 
I want to show that $\mathbb{F}_p(a)$ contains all the roots of $f$. 
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If $a\in \mathbb{F}_p$ then $a^p=a$ and then 
$a^{p^n}=(a^p)^{p^n-1}=a^{p^n-1}=(a^p)^{p^n-2}=a^{p^n-2}=\ldots =a$ 
So, $f(a)=a^{p^n}-a-1=-1\neq 0$. 
Thus it must be $a\notin \mathbb{F}_p$. 
Could you give me a hint how we could show that $\mathbb{F}_p(a)$ contains all the roots of $f$ ?

REPLY [1 votes]: Hint: 
For any $\;t\in\Bbb F_p\;$ , we have that
$$(a+t)^{p^n}-(a+t)-1=a^{p^n}-a-1+t^{p^n}-t=0$$
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