Mod of $n!$

$100!^{100!} = 50!^{100!}\left(51\cdot 52\cdots 99\cdot 100\right)^{100!} = \left(50!^{50!}\right)^{51\cdot52\cdots 99\cdot 100}\left(51\cdot52\cdots 99\cdot 100\right)^{100!} \equiv 0\bmod 50!^{50!}$

$100! = 50! * 51 *52 * \cdots * 99 *100$ So $100!$ is divisible by $50!$

{ 100! }^{ m }={ (50!*51*\cdots *99*100) }^{ m }={ 50! }^{ m }*{ \left( 51*\cdots *99*100 \right) }^{ m } So it is always divisible by ${ 50! }^{ n }$ , if $m\ge n$ .

Here is an example, I hope that you will understand this.

5! = $1 \times 2 \times \boxed{3} \times 4 \times 5$

So the number is divisible by 3

5! mod 3 = 0