Divisibility of Elements of Geometric Sequence from One where First Element is Prime

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

Let $a_0 = 1$.

Let $a_1$ be a prime number.

Then the only divisors of $a_n$ are $a_j$ for $j \in \left\{{1, 2, \ldots, n}\right\}$.

Proof
From Form of Geometric Progression of Integers from One, the elements of $Q_n$ are given by:
 * $Q_n = \left({1, a, a^2, \ldots, a^n}\right)$

From Elements of Geometric Progression from One which Divide Later Elements, each of $a_j$ for $j \in \left\{{1, 2, \ldots, n}\right\}$ are divisors of $a_n$.

From Divisors of Power of Prime, these are the only divisors of $a_n$.