Integer Less One divides Power Less One

Theorem
Let $q, n \in \Z_{>0}$.

Then:
 * $\paren {q - 1} \divides \paren {q^n - 1}$

where $\divides$ denotes divisibility.

Proof
From sum of the geometric progression:


 * $\displaystyle \frac {q^n - 1} {q - 1} = \sum_{k \mathop = 0}^{n-1} q^k$

That is:
 * $q^n - 1 = r \paren {q - 1}$

where $r = 1 + q + q^2 + \cdots + q^{n - 1}$.

As Integer Addition is Closed and Integer Multiplication is Closed, it follows that $r \in \Z$.

Hence the result by definition of divisor of integer.