Question 174
Prove that there are at least $100!$ ways to partition the number $100!$ into summands from the set $\{1!, 2!, 3!, \ldots, 99!\}$. (Partitions differing in the order of summands are considered the same; any summand can be taken multiple times. We remind that $n! = 1 \cdot 2 \cdot \ldots \cdot n$.)