Form of Geometric Sequence of Integers from One

Theorem
Let $P = \left\langle{a_j}\right\rangle_{1 \mathop \le j \mathop \le n}$ be a geometric progression of length $n$ consisting of integers only.

Let $a_1 = 1$.

Then the $j$th term of $P$ is given by:
 * $a_j = a^{j - 1}$

where:
 * the common ratio of $P$ is $a$
 * $a = a_2$.

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

Proof
From Form of Geometric Progression of Integers, the $j$th term of $P$ is given by:
 * $a_j = k q^{j - 1} p^{n - j}$

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

As $a_1 = 1$ it follows that:
 * $1 = k p^{n - j}$

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

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

Thus:
 * $a_2 = q$

Setting $a = a^2$ yields the result as stated.