Problem: Let there be $100$ numbers $x_1, x_2, \cdots, x_{100}$ where all the terms are $\geq 2.$ Is it possible for $x_1!x_2! \cdots x_{100}!=n!$ for some positive integer $n?$
So far, all my progress is that if we choose an arbitrary number as a factorial first, our second factorial must be contained in our first factorial, be equal to our first factorial, or it contains our first factorial.
For example, to make things clearer, assume our first factorial is $50!.$ If our second factorial is something like $3!,$ then we have $(3!)^2 \cdot 4\cdot 5 \cdots \cdot 50.$ If our second factorial is $50!$ we have $(50!)^2$. If our second factorial is something like $150!$ we will have $(50!)^2 \cdot 51 \cdot 51 \cdots 150.$ However, I'm not really sure on how this helps. May I have some assistance? Thanks in advance.
Best Answer
Let's start with $3!=6$. Then $3!5!=5!6=6!=720$. Then $3!5!719!=719!720=720!$. You can continue this procedure to get as many factorials as you wish, whose product will be a factorial. A rather large factorial, but a factorial nontheless.