First Element of Geometric Sequence that divides Last also divides Second

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 $a_0$ be a divisor of $a_n$.

Then $a_0$ is a divisor of $a_2$.

Proof
By hypothesis, let $a_0$ be a divisor of $a_n$.

Aiming for a contradiction, suppose $a_0$ is not a divisor of $a_2$.

From First Element of Geometric Progression not dividing Second it would follow that $a_0$ does not divide $a_n$.

From this contradiction follows the result.