Logic of statement I can see the mathematical implication but could not get the logic, why $5!$ is equal to $^6P_3$? Please help proving why both the expressions are equal without mathematical manipulation!In any case, does there exist any generalization of the above property?
 A: As $\frac{(n!)!}{n!}=(n!-1)!$ for all n , there are infinitely many nontrivial triples (a,b,c) with $\frac{a!}{(a-b)!}=c!$
Choose a=n! , b=n!-n , c=n!-1
A: Richard Guy, Unsolved Problems In Number Theory, 3rd edition, problem B23: Equal products of factorials, begins, 
Suppose that $$n!=a_1!a_2!\dots a_r!,\quad r\ge2,\quad a_1\ge a_2\ge\dots\ge a_r\ge2$$ A trivial example is $$a_1=a_2!\dots a_r!-1,\quad n=a_2!\dots a_r!$$ Dean Hickerson notes that the only nontrivial examples with $n\le410$ are $$9!=7!3!3!2!,\quad10!=7!6!=7!5!3!,\quad16!=14!5!2!$$ and asks if there are any others. Jeffrey Shallit and Michael Easter have extended the search to $n=18160$ and Chris Caldwell has shown that any other $n$ is greater than $10^6$. 
Elsewhere, Guy cites 
MR1646003 (99i:11025) Passow, Eli, On the solutions of $n!=m!r!$,  Far East J. Math. Sci. 6 (1998), no. 3, 337–344. According to the review by Nikos Tzanakis, the author proves that if $r,j,k$ are positive integers, then $$(r+j+k)!=(r+j)!r!$$ has no solution with $j\le1914$ other than $(r,j,k)=(6,1,3)$. 
