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 $N! = A!\cdot B!$ ?
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If you want $N! = A! + B!$, then $A,B <N$. Hence, $N! = A! + B! \leq 2(N-1)!$. This is possible only if $N =2$.
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.
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.
$A!B! = (N!-A!)(N!-B!)$
divide both sides by A!B!