First Element of Geometric Sequence not dividing Second/Proof 2

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

Let $a_0$ not be a divisor of $a_1$.

Then:
 * $\forall j, k \in \left\{{0, 1, \ldots, n}\right\}, j \ne k: a_j \nmid a_k$

That is, if the initial term of $P$ does not divide the second, no term of $P$ divides any other term of $P$.

Proof
From Form of Geometric Progression of Integers, the terms of $P$ are in the form:
 * $(1): \quad: a_j = k q^j p^{n - j}$

where $p \perp q$.

Let $a_0 \nmid a_1$.

That is:

Aiming for a contradiction, suppose that:
 * $\exists i, j \in \left\{{0, 1, \ldots, n}\right\}, i \ne j: a_i \mathop \backslash a_j$

WLOG let $i < j$.

But we have that $p \perp q$.

From Powers of Coprime Numbers are Coprime:
 * $q^{j - i} \perp p^{j - i}$

This can happen only when $q = 1$.

This is incompatible with $q \nmid p$.

From this contradiction, the result follows.