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Let $q=p^k$ with $p$ a prime number. Let the integer $c=\frac{q^n}{1+n(q^n-1)}$ (it's an integer because of Hamming identity) be a power of $p$ and $c\equiv 1 \pmod{q-1}$.

I have to prove that the integer $c$ is a power of $q$.

Considering the hypothesis I have the fact that $c=p^v$ with $0\le v\le k$. So $p^v \equiv 1 \pmod{q-1}$. I don't have other ideas to continue.

Thanks in advance !

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    If you really have $0 \le v \le k$, then $1 \le c \le q$. So then since $c \equiv 1 \pmod{q-1}$, $c$ is either equal to $1$ or $q$, either of which is power of $q$. The question seems rather odd as the number of choices of $n$ for which $c$ is an integer (let alone a power of $p$) is very limited. – 2017-01-18
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    But isn't it obvious that we must have $c\equiv 1 \bmod q-1$? – 2017-01-18

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Since $q\geq 2$ we have that $\frac{q^n}{1+n(q^n-1)}$ is less than $1$ for all $n\geq 2$.

So we must only check the case $n=1$ which always yields $1$.

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    It's Hamming equality, $c\in \mathbb{F}_{q}^{n}$ and by formula for a one-perfect code : $c(1+n(q^n-1))=q^n=(p^k)^n$ – 2017-01-18