First Element of Geometric Sequence not dividing Second/Proof 1

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

Let $a_1$ not be a divisor of $a_2$.

Then:
 * $\forall j, k \in \left\{{1, 2, \ldots, n}\right\}, j \ne k: a_j \nmid a_k$

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

Proof
Let $P_a = \left({a_1, a_2, \ldots, a_n}\right)$ be a geometric progression of natural numbers such that $a_1 \nmid a_2$.

Aiming for a contradiction, suppose $a_1 \mathop \backslash a_k$ for some $k: 3 \le k \le n$.

Let $b_1, b_2, \ldots, b_k$ be the least natural numbers which have the same common ratio as $a_1, a_2, \ldots, a_k$.

These can be found by means of.

From
 * $a_1 : a_k = b_1 : b_k$

Also:
 * $a_1 : a_2 = b_1 : b_2$

and so as $a_1 \nmid a_2$ it follows by :
 * $b_1 \nmid b_2$

From One Divides all Integers it follows that:
 * $b_1 \ne 1$

From :
 * $b_1 \perp b_k$

But as:
 * $a_1 : a_k = b_1 : b_k$

it follows that
 * $a_1 \nmid a_k$

Now suppose $a_j \mathop \backslash a_k$ such that $1 < j < k$.

Let $b_j, \ldots, b_k$ be the least natural numbers which have the same common ratio as $a_j, \ldots, a_k$.

These can be found by means of.

From
 * $a_j : a_k = b_j : b_k$

The other cases follow similarly.