Question: Show that $X^{n} - 1$ has at most $p$ solutions in $\mathbb{Q}_{p}$, and has exactly $p - 1$ solutions if $n = p - 1$. Conclude that if $q < p$ and $(q , p) \neq (2, 3)$ then $\mathbb{Q}_{q} \not\cong \mathbb{Q}_{p}$.
My attempt: From Proposition II.2.4 from Neukrich, we know that $\mathbb{Z}_{p} / p^{n} \mathbb{Z}_{p} \cong \mathbb{Z} / p^{n} \mathbb{Z}$. Furthermore, we have proven earlier that the roots of $X^{n} - 1$ will be in $\mathbb{Z}_{p}$, since $a^{n} = 1 \in \mathbb{Q}_{p} \implies a \in \mathbb{Z}_{p}$.
If we suppose that $n = p - 1$, we can consider $F = X^{p - 1} - 1 \in \mathbb{Z}_{p}[X]$. Notice by Fermat's Little Theorem that $a^{p} \equiv a \mod p$ for all $a \in \mathbb{F}_{p}$. Since the roots of $F$ are non-zero, it follows that $a^{p-1} \equiv 1 \mod p$, or equivalently $a^{p - 1} - 1 \equiv 0 \mod p$. Since the unit group $\mathbb{F}_{p}^{*}$ is of order $p - 1$, it follows that all the elements in $\mathbb{F}_{p}^{*}$ are the roots of $F$. Furthermore, notice that $F' = (p - 1)X^{p - 2} = -X^{p - 2}$ and that $-a^{p-2} \not\equiv 0 \mod p$. Therefore, there are no double roots for $F$. By Hensel's Lemma, it follows that all $p - 1$ roots correspond uniquely with a root of $F$ in $\mathbb{Z}_{p}$. So there are exactly $p - 1$ solutions.
My question is firstly if I did this case right. Furthermore, I would like to know how I could show that $X^{n} - 1$ has at most $p$ solutions in $\mathbb{Q}_{p}$, if anybody has a hint how I could start that I would appreciate that a lot!
