Geometric Sequence in Lowest Terms has Coprime Extremes/Proof 1

Proof
Let $a_0, a_1, a_2, \ldots, a_n$ be natural numbers.

Let $\sequence {a_k}_{0 \mathop \le k \mathop \le n}$ be a geometric progression with common ratio $r$.

Let $a_0, a_1, \ldots, a_n$ be the smallest such natural numbers.

From, let $d_0, d_1$ be the smallest natural numbers such that $d_1 = r d_0$.

From one can build a sequence of $3, 4, \ldots, n$ terms with the same property.

Let the geometric progression so constructed with $n$ terms be $\sequence {b_k}_{0 \mathop \le k \mathop \le n}$.

From, $d_0$ and $d_1$ are coprime.

From, each of the extreme terms of the intermediate geometric progressions is a power of a natural number.

From, each of those extreme terms is coprime.

We have that $a_0, \ldots, a_n$ are the smallest such natural numbers such that $\sequence {a_k}_{0 \mathop \le k \mathop \le n}$ is a geometric progression with common ratio $r$.

We have also constructed $b_0, \ldots, b_n$ to have the same property.

Thus:
 * $a_0 = b_0, a_1 = b_1, \ldots, a_n = b_n$

and the result follows.