Quote:
Originally Posted by henryzz
Just realised that there is an infinite number of examples where (n!)!=n!*(n!1)!
Are there any more apart from this trivial form?

well one condition on this occuring is that one of the factorials involved has to involve the last prime below the factorial being divided by other factorials so for example the condition for example the reason 10!=6!*7! works is because one of the factorials involves 7 a prime number no solution can leave this out. also a key part is that the other factorial has to include possible factorizations of the rest so 8*9*10 = 8*90 = (2*4)*(5*6*3) which means it can barely fall below sqrt(number) when trying to find solutions to number! okay I made an error but I was trying to limit the cases.