I'm trying to prove that $X^n-1$ has no repeated irreducible factors over $\mathbb{F}_q$ if and only if $gcd(q,n)=1$. So far I've completed one implication: If $gcd(q,n) \neq 1$, then $n=p^em$ for some $e>0, p \nmid m$. The Frobenius yields $X^n-1=X^{mp^e}-1=(X^m-1)^{p^e}$ which obviously has repeated irreducible factors.
However, I'm stuck on the other implication. I know that $f \in \mathbb{F}_q[X]$ has no repeated factors if and only if $gcd(f,f')=1$ where $f'$ is the formal derivative of $f$. But I do not know how I could apply this for $X^n-1$ and its formal derivative $nX^{n-1}$. Any hint would be highly appreciated.
Suppose $\gcd(n,p)=1$.
Let $I$ be the ideal of $F_q[X]$ generated by $X^n-1,nX^{n-1}$.
From $\gcd(n,p)=1$, it follows that $n$, when regarded as an element of $F_q$, is a unit, hence \begin{align*} & nX^{n-1}\in I \\[4pt] \implies\;& X^{n-1}\in I \\[4pt] \implies\;& X^n\in I \\[4pt] \implies\;& X^n-(X^n-1)\in I \\[4pt] \implies\;& 1\in I \\[4pt] \implies\;& I=(1) \\[4pt] \end{align*} so in $F_q[X]$, we have $\gcd(X^n-1,nX^{n-1})=1$.
Here's another way . . .
Suppose $\gcd(n,p)=1$, and suppose $X^n-1$ and $nX^{n-1}$ have a nonunit common factor in $F_q[X]$.
Then $n$ is a unit, when regarded as an element of $F_q$, and $n > 1$, when regarded as an element of $\mathbb{Z}$.
But then since $F_q[X]$ is a UFD, it follows that $X^n-1$ and $nX^{n-1}$ must have a common prime factor, contradiction, since the only prime factor of $nX^{n-1}$ is $X$, which is not a factor of $X^n-1$.