Unique representation: $a!cdot b!cdot c!=m!cdot n!cdot p!$

Combine $6!7!=10!$ with @N.S.'s answer

$$6!7!(n!)!=10!n!(n!-1)!$$

• $6!7!24!=4!10!23!$
• $6!7!120!=5!10!119!$
• $6!7!720!=6!10!719!$

The famous $$n! cdot (n!-1)!=(n!)!=(n!)! cdot 1! $$ is a counterexample for 2, and can be extended to $3$:

$$1!cdot1!cdot((n!)!)!=(n!)! ((n!)!-1)!=n! cdot(n!-1)! cdot((n!)!-1)!$$

$4!4!25! = 5!5!24!$

$5!5!36! = 6!6!35!$

$6!4!35! = 7!5!34!$ ....

$a!b![(a+1)(b+1)]! = (a+1)!(b+1)![(a+1)(b+1)-1]!$

The part about two is wrong

$$ 4! 15! = 7! 13! $$

$$ 3! 20! = 5! 19! $$

$$ 4! 30! = 6! 29! $$

$$ 5! 42! = 7! 41! $$

$$ 6! 56! = 8! 55! $$

$$ 18! 57! = 22! 54! $$

$$ 7! 66! = 14! 62! $$

$$ 7! 72! = 9! 71! $$

$$ 8! 90! = 10! 89! $$

=========================================

$$ 3! 3! 16! = 4! 4! 15! = 4! 7! 13! = 2^{17} cdot 3^8 cdot 5^3 cdot 7^2 cdot 11 cdot 13 $$ $$ 3! 8! 13! = 4! 6! 14! = 2^{18} cdot 3^8 cdot 5^3 cdot 7^2 cdot 11 cdot 13 $$

$$ 3! 4! 20! = 4! 5! 19! = 2^{18} cdot 3^8 cdot 5^3 cdot 7^2 cdot 11 cdot 13 $$