Question:Prove that $x^{p^n} -x +1$ is irreducible over $\mathbb{F_p}$ only when n=1 or n=p=2
I know it is a duplicate question. However, someone gave some nice hints on this problem and I want to follow the hints he gave.
Hint (1): Note that if $\alpha$ is a root, then so is $\alpha+a$ for any $a \in \mathbb{F_{p^n}}$
Hint (2): Show that this implies that $F_p(\alpha)$ contains $\mathbb{F_{p^n}}$
Hint (3): Show that $[\mathbb{F_{p}(\alpha)}:\mathbb{F_{p^n}}]=p$
Hint(1) and Hint(2) are done:
Hint(1) is easy to prove, just use two properties: $(\alpha+\beta)^p=\alpha^p+\beta^p$ and $\forall a \in \mathbb{F_{p^n}}$, $a$ is a root of $x^{p^n}-x+1$.
Hint(2): Since $F_p(\alpha)$ is a Galois Extension, we have $F_p(\alpha)$ contain all the roots of $x^{p^n} -x +1$, thus contain $\mathbb{F_{p^n}}$.
However, I stuck on Hint(3), and actually Hint(3) is very important, once we get $[\mathbb{F_{p}(\alpha)}:\mathbb{F_{p^n}}]=p$, we will get $[\mathbb{F_{p}(\alpha)}:\mathbb{F_{p}}]=n \cdot p$ ,and we know $[\mathbb{F_{p}(\alpha)}:\mathbb{F_{p}}]=p^n$, thus we can finish the proof.
Can anyone give me any idea about how to prove hint(3)?