Let $p$ be prime and $n = p^e$ where $e \ge 1$. Prove the following:
If $a \equiv 1 \pmod p$ then $1 + a + a^2 + \ldots + a^{n-1} \equiv 0 \pmod n$
Note that the the first congruence is$\pmod p$ and the second is$\pmod n$.
What I tried: I know $1 + a + a^2 + \ldots + a^{n-1} = \Phi_p(a)\cdots\Phi_{p^e}(a)$ where $\Phi_p$ is the $p$th cyclotomic polynomial. I want to show $a$ is a root$\pmod n$ in $\Phi_{p^i}(a)$ for some $i$. I know how to write $\Phi_{p^i}$ in terms of $\Phi_{p}$. However, I don't think $a$ is necessarily a root in any of these.
Best Answer
Hint: write $$1+a+a^2+\dots+a^{n-1}=\frac{1-a^{(p^e)}}{1-a}=\prod_{k=0}^{e-1}\frac{1-a^{(p^{k+1})}}{1-a^{(p^k)}}$$
Now you only need to prove that all the $\frac{1-a^{(p^{k+1})}}{1-a^{(p^k)}}$ will be divisible by $p$. (Here I assumed $a\ne1$, but we can verify it directly in the case $a=1$ or we can reduce to the case $a\ne1$ by adding $p^e$ which doesn't change the value mod $p^e$)