algebra precalculus - Why k should be odd?

My teacher once said, for any positive number $\ n,$ $\ n^k - 1$ would always have $\ n-1$ as a factor for all positive odd values of $\ k$. Could anyone tell me the proof? I have written my approach below.

Assuming $n-1$ to be a factor.

$n^k - 1 = (n-1)x$

$n^k = (n-1)x + 1$

$k = \lg( (n-1)x + 1 ) / \lg (n)$

If my approach is right could you tell me where should I go from here?

Edit : This clears my doubt.

Edit 2 : My question is "why for any positive number $\ n,$ $\ n^k + 1$ would always have $\ n+1$ as a factor for all positive odd values of $\ k$ ".

Answer

More generally, if $a,b\in\mathbb Z$, $n\in\mathbb Z^+$, then $a-b\mid a^n-b^n$.

It follows from $$a^n-b^n=(a-b)\left(a^{n-1}+a^{n-2}b+\cdots+b^{n-1}\right)$$

Edit: Since it seems you wanted to ask about the fact that for $n\in\mathbb Z$ with odd $k\in\mathbb Z^+$ we have $n+1\mid n^k+1$:

More generally, for $a,b\in\mathbb Z$ and odd $n\in\mathbb Z^+$ we have $$a^n+b^n=(a+b)\times$$

$$\times \left(a^{n-1}-a^{n-2}b+a^{n-3}b^2-\cdots-ab^{n-2}+b^{n-1}\right)$$