How to factorize $x^{p^2}-1$?

Question

We know $$x^p−1 = (x-1)(x^{p−1}+x^{p−2}+⋯+x+1).$$ The question is how to factorize $x^{p^2}-1$?

I expanded $x^{p^2}-1$ as $(x-1)(x^{p^2-1}+x^{p^2-2}+\cdots+x+1)$ following the pattern, but I failed in reducing further the second expression(which is suggested as "can"). Can somebody help?

Additional context: This is a random question from a group chat of people interested in math. This group has a mixed student body in high school and early undergraduate. I searched and used this resource to remind myself how to correctly factorize $x^p-1$: https://en.wikipedia.org/wiki/Root_of_unity, but I am not able to understand many other contents addressed in this resource such as group, field, etc.

Answer 1

With $y:=x^p$,$$x^{p^2}-1=y^p-1=(y-1)\sum_{k=0}^{p-1}y^k=(x^p-1)\sum_{k=0}^{p-1}x^{kp}=(x-1)\sum_{j=0}^{p-1}x^j\sum_{k=0}^{p-1}x^{kp}.$$

Answer 2

$$(x - 1) \frac{x^p - 1}{x-1} \frac{x^{p^2} - 1}{x^p-1}$$