Question 120
Prove that there exist pairwise distinct positive integers $a_{0}, a_{1}, a_{2}, \ldots, a_{1000}$ such that
$$
a_{0}! = a_{1}!\, a_{2}!\, \ldots\, a_{1000}!
$$
Here $n! = 1 \times 2 \times \cdots \times n$ as usual.
$$
a_{0}! = a_{1}!\, a_{2}!\, \ldots\, a_{1000}!
$$
Here $n! = 1 \times 2 \times \cdots \times n$ as usual.