Factorial puzzle
I just noticed that 10!=6!*7!.
Are there any other instances where x!=y!*z! with y and z > 1? If not can anyone prove this? 
Just realised that there is an infinite number of examples where (n!)!=n!*(n!1)!
Are there any more apart from this trivial form? 
[QUOTE=henryzz;399161]Just realised that there is an infinite number of examples where (n!)!=n!*(n!1)!
Are there any more apart from this trivial form?[/QUOTE] 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. 
[URL="http://oeis.org/A034878"]Numbers n such that n! can be written as the product of smaller factorials.[/URL]

[QUOTE=henryzz;399160]I just noticed that 10!=6!*7!.
Are there any other instances where x!=y!*z! with y and z > 1? If not can anyone prove this?[/QUOTE] No other examples for y,z < 10000 except the trivial cases. 
[QUOTE=ATH;399200]No other examples for y,z < 10000 except the trivial cases.[/QUOTE]
using pari I've checked for solutions using up to the last 10000 factorials before a factorial up to 30000! I think: [CODE]factorfactorial(e=0)={ a=parvector(10000,n,(n+e)!); parfor(x=10,#a, d=precprime(x);xd, c, if(c==0c>=10000,next()); b=setintersect(vector(c1,t,a[x]/a[x+t(c+1)]),a); if(b==[],, print(x","b) ) ); e=e+#a; print(e); factorfactorial(e) };[/CODE] doh I see an error now and it might of been slowing it down. 
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