A number is highly composite if it is the smallest number that has more divisors that any number less than it.
Let $(h_n)_{n\ge 1}$ be the sequence of highly composite numbers and $\omega(n)$ denote the number of distinct prime divisors of $n$. I noticed that $\omega(h_n)$ is a nondecreasing sequence but I don't know why. Here's my attempt,
Assume not then we can find an $n$ such that $$\omega(h_n)>\omega(h_{n+1})$$ Write $$h_n=\prod_{i=1}^rp_i^{a_i}\text{ and } h_{n+1}=\prod_{i=1}^sp_i^{b_i}$$ with $r>s$ and $p_i$ is the ith prime number. Also the $a_i$'s and the $b_i$'s are both nonincreasing sequences (those are all well-known facts about highly composites). Therefore $$1>\frac{h_n}{h_{n+1}}=\frac{\prod_{i=1}^sp_i^{a_i}\prod_{i=s+1}^rp_i^{a_i}}{\prod_{i=1}^sp_i^{b_i}}=\prod_{i=1}^sp_i^{a_i-b_i}\prod_{i=s+1}^rp_i^{a_i}$$ Now I don't know what to do I think if we can somehow get a handle on $a_i-b_i$ then we may get a contradiction.
