Diophantine Factorial Equation

Question

Prove that there exist pairwise distinct positive integers $a_0, a_1, a_2, \ldots, a_{1000}$ such that $a_0! = a_1!a_2! \cdots a_{1000}!.$

Answer 1

Hint: You can let somebody else choose $a_1$ through $a_{999}$, then you can choose $a_0, a_{1000}$ to succeed.