# Finding the maximum $k$ such that $(7!)!$ is divisible by $(7!)^{k!}\cdot(6!)!$

divisibilityfactorialproblem solving

If $$(7!)!$$ is divisible by $$(7!)^{k!}\cdot(6!)!$$, then what is the maximum value of $$k$$?

At first glance I couldnt think of anything except Legendre's formula for calculating powers of a prime in a factorial. It would indeed be cumbersome task of calculating all primes that occur in $$(6!)!$$. So I came up with another possible solution but I don't know if it gives me the maximum value of $$k$$ or not. Do let me know what you think.

If we see this situation as dividing $$7!$$ different objects in $$6!$$ groups and each group having $$7$$ objects each then the number of ways of making these different groups will be $$\dfrac{(7!)!}{((7!)^{6!})((6!)!)}$$ Now, this will give me $$k=6$$, and this ensures that quotient is a natural number, too.

You've already shown that $$(7!)^{6!}((6!)!)$$ divides $$(7!)!$$. If $$(7!)^{7!}((6!)!)$$ divides $$(7!)!$$, then $$(7!)^{7!}$$ divides $$(7!)!$$, so $$(7!)^{7!}\leq (7!)!$$. However, $$n^n>n!$$ for all $$n>1$$.