Condition for Divisibility of Powers of Prime

Theorem
Let $p$ be a prime.

Let $k, l \in \Z_{>0}$.

Then:
 * $p^k \mathop \backslash p^l \iff k \le l$

Necessary Condition
Let $k \le l$.

Then:
 * $l - k \ge 0$

Thus $p^k, p^{l-k} \in \Z$ such that $p^l = p^k p^{l-k}$.

Thus:
 * $p^k \mathop \backslash p^l$

Sufficient Condition
Let $p^k \mathop \backslash p^l$.

Then:
 * $\exists b \in \Z_{>0}: p^l = p^k b$

By the Fundamental Theorem of Arithmetic, $b$ has a unique prime decomposition.

Either $b = 1$ (in which case $k = l$) or $b$ has a prime decomposition consisting entirely of $p$'s.

In this case:
 * $\exists m \in \Z: b = p^m$

Hence:
 * $p^{l-k} = p^m$

Thus from the Fundamental Theorem of Arithmetic:
 * $l - k = m > 0$

Thus:
 * $l > k$

The result follows from combining the two results.