$F_p(a)$ contains all the roots of $f$ 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 $f$ ?

Let $|F_p(a)|=g$. Then as $F_p(a)^\times$ is a group of order $g-1$, all elements of $F_p(a)$ satisfy $x^g-x=0$. Because $F_p$ is a field, this polynomial, which has all distinct roots as its derivative is identically $-1$ has distinct roots in the algebraic closure, in particular, all elements of $F_p(a)$ are roots of it. However, since--in particular--$a\in F_p(a)$ the minimal polynomial for $a$ divides $x^g-x$. So since all the roots of the minimal polynomial of $a$ are roots of $x^g-x$, a fortiori all the roots of the minimal polynomial of $a$ are in $F_p(a)$.
