Let $f \in \mathbb{F}_{q}[x]$, and let $x^n - 1$ divide $f^{k}$ in $\mathbb{F}_q[x]$, for some natural number $k$.
If $gcd(n,q) = 1$, then apparently I can deduce that $x^n - 1$ divides $f$?
How is this true? Apparently it is enough to check that the polynomial $x^n - 1$ is square-free, which is easy to see, but I don't understand how this shows that $x^n - 1$ divides $f$?
Assume $n$ and $q$ are coprime. Then $x^n-1$ is coprime with its derivative, thus is the product of its irreducible factors.
If $p$ is an irreducible factor of $x^n-1$, then $p|f^k$. Thus $p$ must divide $f$.
Then the product of the irreducible factors of $x^n-1$, which is $x^n-1$, divides $f$.