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

 A: 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 $n>1$.
A: ETA The fraction in the OP is indeed integral, I just wasn't able to see what was happening combinatorially before. In any event this is an alternative approach:
To see this, first write
$$(7!)! =$$ $$(7!)((7!-1)\cdots(6!+1))×(6!)!.$$
Then from this, write
$$A=(7!)(7!-1)\cdots(6!+1).$$
We first show that $k$ is at least $6$. The crux is showing that $k$ is at least $6$ is showing that $(7!)^{6!}$ divides $A$.
Now it isn't too hard to see that $9^{6!}$ divides $A$, and so does $5^{6!}$ [because there are at least $\frac{6×6!}{3}\ge 2×6!$ and $\frac{6×6!}{5}$ integers $l$ in the interval $[6!+1,7×6!]$ that are multiples of $3$ and $5$ respectively]. Being a bit more careful, one can see that $(16)^{6!}$ also divides $A$. So from all this it follows that $(9×5×16)^{k!}$ $=$ $6!^{6!}$ divides $A$. So indeed, to show that $(7!)^{6!}$ divides $A$, it suffices to show that $7^{6!}$ divides $A$. We do this next.
Let us find the largest power of $7$ that divides $A$. The largest integer $K$ such that $7^K$ divides $A$ satisfies
$$K=K_1+K_2+K_3+K_4,$$ where for each $i=1,2,3,4$, the integer $K_i$ are the number of multiples of $7^i$ in the interval $[6!+1,7!]$ $=[721,5040]$. [Note that $7^q > 5040$ for all $q \ge 5$.] Now, $K_4=2$, $K_3=12$, $K_2=88$, and $K_1=618$. So
$$K=K_1+K_2+K_3+K_4$$
$$=618+88+12+2 = 720=6!.$$
So the largest integer $K$ s.t. $7^K$ divides $A$ is exactly $K=6!$. Down to the wire! Thus indeed $7^{6!}$ does divide $A$, and as noted above, this suffices to show that $(7!)^{6!}$ divides $(7!)!$. And so $k$ as in the original problem is at least $6$.
As noted in the answer above, $k$ must be smaller than $7$ so $k$ must be exactly $6$, and the result follows. $\surd$

That the largest integer $K$ such that $7^K$ divides $A$, satisfies $K=720$ exactly, implies not only that $6!$ is the largest factorial such that $(7!)^{6!}×(6!)!$ divides $(7!)!$, it also shows the stronger statement that $K=6!$ is the largest integer such that $(7!)^K×(6!)!$ divides $(7!)!$.
