Factorial expressed in terms of two other factorials Can the factorial of $N$ always be expressed by the sum(addition and subtraction) or the product of two other factorials?
Do there always exist integer $A$ and $B$ such that $N! = A! + B!$, or $N! = A! - B!$, or $N! = A!\cdot B!$ ?
 A: If you want $N! = A! + B!$, then $A,B <N$. Hence, $N! = A! + B! \leq 2(N-1)!$. This is possible only if $N =2$.
A: Multiplication has been asked on this site before. The general example is
$$  (n! - 1)! \cdot n! = (n!)!   $$ with examples such as
$$  n=3; \; \; \; 5! \cdot 6 = 6!  $$ 
$$  n=4; \; \; \; 23! \cdot 24 = 24!  $$
$$  n=5; \; \; \; 119! \cdot 120 = 120!  $$
The only known nontrivial example is 
$$  6! \cdot 7! = 10!   $$
Well, maybe I will use capital letters for this. If $K! \cdot M! = N!$ and $K<M<N,$ we know that $N$ cannot be a prime, indeed there cannot be a prime $p$ with $M+1 \leq p \leq N.$ So the size of prime gaps is part of the discussion of possible other nontrivial examples.
A: No, in fact it is rare.  Any factorial above $2!$ is more than a factor of $2$ away from any other factorial, which eliminates addition and subtraction.  Consideration of how many factors of $2$ are in each factorial eliminates most of the others aside from $N!=0!N!=N!1!$.  All examples are within $0!,1!,2!$.  We have $2!=1!+1!=1!+0!=0!+1!=0!+0!$ for all the additions and you can derive the subtractions from there.  For multiplication, $2!=0!2!=1!2!, 1!=1!1!=0!0!$ and the obvious others.
A: If $N!=A!+B!$
then
$A!=N!-B!$
and
$B!=N!-A!$
so
$A!B! = (N!-A!)(N!-B!)$
divide both sides by A!B!
$1=(N!/A!-1)(N!/B!-1)$
