$(n-1)^2$ divides $(n^k -1) $ if and only if $(n-1)$ divides $k$ where $n\geq 2$ and $ n \geq 1$
$$n^k-1 = (n-1)(n^{k-1} + n^{k-2} + \ldots + 2 + 1)$$
So $$(n-1)^2 \;\big|\;(n^k-1) \iff (n-1) \; \big| \; (n^{k-1} + n^{k-2} + \ldots + 2 + 1)$$
So I want to prove $(n-1) \; \big| \; (n^{k-1} + n^{k-2} + \ldots + 2 + 1) $ if and only if $ (n-1) \; \big| \; k$
How can I proceed from here?
