Irreducibility of $x^{q-1}-\gamma$ in finite field $\mathbb{F}_q$ Let $\gamma$ be a primitive element of $\mathbb{F}_q$.

How to prove that polynomial $x^{q-1}-\gamma$ is irreducible over $\mathbb{F}_q$?

Denote $\alpha$ as a root of  $x^{q-1}-\gamma$, I want to show that $[\mathbb{F}_q(\alpha):\mathbb{F}_q]$is $q-1$. One can see $\alpha$ has order $(q-1)^2$, but how does this related to degree of field extension?
 A: Let $n=[\mathbb{F}_q(\alpha):\mathbb{F}_q]$.

Since $\alpha$ is a root of $x^{q-1}-\gamma$, it follows that $n\le q-1$.

Our goal is to show that $n=q-1$.

Let $s=o(\alpha)$.

As you observed (and it's a key observation), from $o(\gamma)=q-1$ and $\alpha^{q-1}=\gamma$, we get $\alpha^{(q-1)^2}=1$, so $s{\,\mid\,}(q-1)^2$.

But in fact, $s=(q-1)^2$, as we now show . . .
\begin{align*}
&
\alpha^s=1
\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\;\;\;\;
\\[4pt]
\implies\;&
\alpha^{(q-1)s}=1
\\[4pt]
\implies\;&
\gamma^s=1
\\[4pt]
\implies\;&
(q-1){\,\mid\,}s
\\[4pt]
\implies\;&
s=(q-1)t\;\,\text{for some positive integer $t$}
\\[4pt]
\implies\;&
\alpha^{(q-1)t}=1
\\[4pt]
\implies\;&
\gamma^t=1
\\[4pt]
\implies\;&
(q-1){\,\mid\,}t
\\[4pt]
\implies\;&
(q-1)^2{\,\mid\,}s
\\[4pt]
\implies\;&
s=(q-1)^2
\\[4pt]
\end{align*}
as claimed.

From $[\mathbb{F}_q(\alpha):\mathbb{F}_q]=n$ we get $|\mathbb{F}_q(\alpha)|=q^n$, hence $o(\alpha){\,\mid\,}(q^n-1)$.
\begin{align*}
\text{Then}\;\;&
o(\alpha){\,\mid\,}(q^n-1)
\\[4pt]
\implies\;&
(q-1)^2{\,\mid\,}(q^n-1)
\\[4pt]
\implies\;&
(q-1){\Large{\,\mid\,}}\frac{q^n-1}{q-1}
\\[4pt]
\implies\;&
(q-1){\Large{\,\mid\,}}\sum_{k=0}^{n-1}q^k
\\[4pt]
\implies\;&
\sum_{k=0}^{n-1}q^k\equiv 0\;\bigl(\text{mod}\;(q-1)\bigr)
\\[4pt]
\implies\;&
\sum_{k=0}^{n-1}1^k\equiv 0\;\bigl(\text{mod}\;(q-1)\bigr)
\qquad
\Bigl(\text{since $q\equiv 1\;\bigl(\text{mod}\;(q-1)\bigr)$}\Bigr)
\\[4pt]
\implies\;&
n\equiv 0\;\bigl(\text{mod}\;(q-1)\bigr)
\\[4pt]
\implies\;&
(q-1){\,\mid\,}n
\\[4pt]
\implies\;&
n\ge q-1
\\[4pt]
\implies\;&
n=q-1
\qquad
\Bigl(\text{since we already have $n\le q-1$}\Bigr)
\\[4pt]
\end{align*}
as was to be shown.
