Showing that $\dfrac{q^n -1}{(q-1)\gcd(n,q-1)}$ is an integer for $q = p^k$, $p$ prime.

Question

Does anyone know how to prove that $\dfrac{q^n -1}{(q-1)\gcd(n,q-1)}$ is an integer for $q = p^k$, $p$ prime using only basic algebra?

It is possible to show this by showing that there is a subgroup $C$ of $\mathbb{F}_{q^n}^*$ such that $ \#C = (q-1)\gcd(n,q-1)$, but is it possible to find a simpler proof?

Answer 1

In modulo $\gcd(n,q-1)$ we have:

\begin{align*} \frac{q^n-1}{q-1} & = q^{n-1} + \cdots + q + 1 \\ & = 1 + \cdots + 1 + 1 \\ & \equiv n \ \equiv \ 0 \end{align*}