Form of Geometric Sequence of Integers from One

Theorem
Let $Q_n = \left\langle{a_j}\right\rangle_{0 \mathop \le j \mathop \le n}$ be a geometric progression consisting of integers only.

Let $a_0 = 1$.

Then the $j$th term of $Q_n$ is given by:
 * $a_j = a^j$

where:
 * the common ratio of $Q_n$ is $a$
 * $a = a_1$.

Thus:
 * $Q_n = \left({1, a, a^2, \ldots, a^n}\right)$

Proof
From Form of Geometric Progression of Integers, the $j$th term of $Q_n$ is given by:
 * $a_j = k q^j p^{n - j}$

where:
 * the common ratio of $Q_n$ expressed in canonical form is $\dfrac q p$
 * $k$ is an integer.

As $a_0 = 1$ it follows that:
 * $1 = k p^{n - j}$

from which it follows that:
 * $k = 1$
 * $p = 1$

and the common ratio of $Q_n$ is $q$.

Thus:
 * $a_1 = q$

Setting $a = a_1$ yields the result as stated.