Prime Decomposition of Highly Composite Number

Theorem
Let $n$ be a highly composite number.

Let the prime decomposition of $n$ be expressed as:
 * $n = \ds \prod_{k \mathop \in \N} {p_k}^{r_k}$

where $p_k$ denotes the $k$th prime.

Then the sequence $\sequence {r_k}$ is decreasing.

That is:
 * $\forall k \in \N: r_k \ge r_{k + 1}$

Proof
Let $n = \ds \prod_{k \mathop \in \N} {p_k}^{r_k}$ be highly composite.

By definition of divisor count function:
 * $\map {\sigma_0} n = \ds \prod_{k \mathop \in \N} \paren {r_k + 1}$

$r_{l + 1} > r_l$ for some $l \in \N$.

Consider $m \in \Z$ whose prime decomposition of $n$ is expressed as:
 * $m = \ds \prod_{k \mathop \in \N} {p_k}^{s_k}$

where:
 * $\forall j < l: s_j = r_j$
 * $\forall j > l + 1: s_j = r_j$
 * $s_l = r_{l + 1}$
 * $s_{l + 1} = r_l$

We have that:

Thus:
 * $\map {\sigma_0} m = \map {\sigma_0} n$

Now we have that:
 * $r_l < r_{l + 1}$

and so:
 * $s_l > s_{l + 1}$

and so:

Then:
 * $p_{l + 1} > p_l$

and so:
 * ${p_{l + 1} }^c > {p_l}^c$

from which it follows that:
 * $m < n$

while:
 * $\map {\sigma_0} m = \map {\sigma_0} n$

But $n$ is highly composite.

This means that if $\map {\sigma_0} m = \map {\sigma_0} n$, then $n \le m$.

This is contradicted by $n > m$.

Thus by Proof by Contradiction:
 * $\forall k \in \N: r_k \ge r_{k + 1}$