$g^{0p^nk}+g^{p^nk}+g^{2p^nk}+...+g^{(p-2)p^nk} \equiv 0$ (mod $p^n)$ if $p-1$ doesn't divide $k$

Question

Let $p$ be prime and $n\geq 1$. Let $g$ be the integer equivalent of some generator for $(\mathbb{Z}/p\mathbb{Z})^ \times$.

Let $k\in \{0,1,2,...\} \subseteq \mathbb{Z}$ such that $p-1$ does NOT divide $k$.

I want to show that $g^{0p^nk}+g^{p^nk}+g^{2p^nk}+...+g^{(p-2)p^nk} \equiv 0$ (mod $p^n)$

I tried writing the sum as $g^{0p^nk}+g^{p^nk}+g^{2p^nk}+...+g^{(p-2)p^nk}=\frac {g^{(p-1)p^nk}-1}{g^{p^nk}-1}$ but I don't see why this would have to be divisible by $p^n$

Answer 1

You are almost done with the formula for the geometric sum as fraction which you write. Just remember that:

1. The multiplicative group $(\Bbb Z/p)^\times$ ist cyclic of order $p-1$. So $g^{(p-1)k} \equiv 1$ mod $p$.

2. On the other hand, the condition that $k$ does not divide $p-1$ implies that $g^{p^n k} \equiv g^k \not \equiv 1$ mod $p$, in other words, the denominator of your fraction is not divisible by $p$.

3. As you just asked in $a \equiv b$ (mod $p$) implies $a^{p^n} \equiv b^{p^n}$ (mod $p^n$)?, the congruence from 1. further implies that $g^{(p-1)k p^n} \equiv 1^{p^n} = 1$ mod $p^n$. In other words, the numerator of your fraction is ...