Prove that $x-1$ divides $x^n-1$ In algebra & polynomials, how do we prove that
$$x-1 \mid x^n -1?$$

Factorization method
Clearly, $x^{n} - 1 = (x - 1)(x^{n - 1} + x^{n - 2} + \dots + 1)$.  Thus, $(x - 1) | (x^{n} - 1)$ since there exists the polynomial namely $(x^{n - 1} + x^{n - 2} + \dots + 1)$ that is multiplied by $(x - 1)$ to obtain $(x^{n} - 1)$.
Substitution method
Suppose that $x - 1 = 0 \rightarrow x = 1$.  If we substitute that value for $(x^{n} - 1)$, then clearly $1^{n} - 1 = 0$.  This shows that $(x^{n} - 1)$ has a factor $(x - 1)$.
