Condition for Divisibility of Powers of Prime

Theorem
Let $$p$$ be a prime and let $$k, l \in \Z_+$$.

Then $$p^k \backslash p^l \iff k \le l$$.

Proof

 * 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 $$p^k \backslash p^l$$.


 * Let $$p^k \backslash p^l$$.

Then $$\exists b \in \Z_+: p^l = p^k b$$

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

Either $$b = 1$$ (in which case $$k - l$$) or have 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 cases.