Show that $\sigma(p^n)=\frac{p^{n+1}-1}{p-1}$ , where $p$ is prime.
I know I can't take for granted for just plugging a question and request for the solution. But I have tried it by using the proof of $\sigma(n)$ but somehow my solution is not convincing, and I think I am not in the right path. Can anyone please guide me? Will appreciate it alot...
This is the formula for $\sigma(n)$ :
You don't need anything to prove it. If $p$ is a prime, the only prime divisor of $p^n$ is $p$. Also, for all natural $k \leqslant n$, we have $p^k | p^n$, and these are the only divisors of $p^n$ since $p$ is a prime.
Thus, $$\sigma(p^n)=1+p+p^2+\cdots+p^n = \frac{p^{n+1} - 1}{p-1}$$