Prove that $(x^n - 1)$ can divide $(x^{kn} - 1)$ without any remainder. I would like your help with proving that $(x^n - 1)$ can divide $(x^{kn} - 1)$ without any remainder.
I understand that both of these functions can be translated to a similar form such as $(x^2-1)=(x-1)(x+1)$.
But I'm not really sure how to do so.
Thank you.

Notice that $x^{kn} - 1 = (x^n)^k - 1.$ This implies that $x^{kn} - 1 = (x^n - 1)P(x).$
