TITLE: Prove that $x-1$ is a factor of $x^n-1$
QUESTION [9 upvotes]: Prove that $x-1$ is a factor of $x^n-1$.

My problem: I already proved it by factor theorem† and by simply dividing them. I need another approach to prove it. Is there any other third approach to prove it? If yes please share it. I will be very thankful to you.
Thanks.

†: Factor Theorem: $x -r$ is a factor of $f(x)$ if and only if $f(r) = 0$.

REPLY [6 votes]: Since the question is tagged "abstract algebra" let's use a little, viz. congruences.
Proof $\,\ {\rm mod}\,\ x-1\!:\,\ x\equiv 1\,\color{#c00}{\overset{\rm CP}\Rightarrow}\, x^n\equiv 1^n \ $ thus $\ x-1\mid x^n-1$
using $\,\rm\color{#c00}{CP}$ = Congruence  Power Rule (or iterated Product Rule), whose simple proof is exactly the same as it is for the ring of integers, since it uses only commutative ring laws.

Or, specializing $\,a = x\!-\!1\,$ below (an inductive proof of first term of a binomial expansion), immediately yields that $\,x^n = 1 + (x\!-\!1)k,\,$ for $\,k\in\Bbb Z[x],\ $ so $\ x\!-1\mid x^n\!-\!1\,$ in $\,\Bbb Z[x].$
$$\begin{align} (1+ a)^n\, \ \  =&\,\ \ 1 + ak\qquad\qquad\quad {\rm i.e.}\ \ P(n)\\[1pt]
\Rightarrow\ (1+a)^{\color{#c00}{n+1}}\! =&\  (1+ak)(1 + a)\\[2pt]  
 =&\,\ \ 1+ a\,(\underbrace{k\!+\!1\!+\!ka\!}_{\large k'})\ \ \ {\rm i.e.}\ \ P(\color{#c00}{n\!+\!1})\\  
  \end{align}\qquad$$
The above is essentially a special case of the prior proof using congruences, without using the language of congruences.