Divisors of Power of Prime

Theorem
Let $p$ be a prime number.

Let $n \in \Z_{> 0}$ be a (strictly) positive integer.

Then the only divisors of $p^n$ are $1, p, p^2, \ldots, p^{n - 1}, p^n$.

Proof
First it is necessary to establish that every element of the set $\left\{{1, p, p^2, \ldots, p^{n - 1}, p^n}\right\}$ is in fact a divisor of $p^n$.

For any $j \in \left\{{1, 2, \ldots, n}\right\}$ we have that
 * $p^n = p^j p^{n - j}$

and so each of $1, p, p^2, \ldots, p^{n - 1}, p^n$ is a divisor of $p^n$

Let $a \in \Z_{>0}: a \notin \left\{{1, p, p^2, \ldots, p^{n - 1}, p^n}\right\}$.

Let $a = p^j$ where $j \in \Z: j > n$.

Then:
 * $p^j = p^n p^{j - n}$

and so $p^n$ is a divisor of $p^j$.

Hence $p_j \nmid p^n$.

Now let $a \notin \left\{{p^k: k \in \Z_{>0} }\right\}$.

Then:
 * $\exists q \in \Bbb P: q \mathop \backslash a$

where:
 * $\Bbb P$ is the set of all prime numbers
 * $\backslash$ denotes divisibility.

Suppose $a \mathop \backslash p^n$.

From Divisor Relation is Transitive it follows that $q \mathop \backslash p^n$.

From Euclid's Lemma for Prime Divisors: General Result it follows that:
 * $q \mathop \backslash p$

As $p$ is a prime, by definition its only divisors are $1$ and $p$.

This contradicts the supposition that $q$ is a divisor of $p^n$.

Hence $a \nmid p^n$.