IZhO 2019 Problem 1

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 $n!=1 \cdot 2 \cdot \ldots \cdot n$.)