Oblique Tides

We need to find $j$ such that $\frac{(7!)!}{(6!)!\times(7!)^{j!}}$ is an integer.

We can try to find the similarity between this expression and the expression for finding the ways of grouping things.

For example, to divide 10 things into 5 groups with 2 things each, the number of ways to divide them is given by $\frac{10!}{(2!)^{5} \times (5!)}$ .
Let us get back to this question. The given expression looks similar to the number of ways to divide $7!$ things into $6!$ groups of $7$ things each which is $\frac{(7!)!}{(7!)^{6!}\times(6!)!}$ . Since it is the number of ways to divide something, it must be an integer. Thus we can see $j=6$ satisfies the divisibility property.

BUT how to make sure that $j=7$ and above doesn't satisfy the criteria? Well to solve that problem, we can see that for any positive integer $a$ , $a^{a}$ is always greater than $a!$ as multiplying $a$ over and over again will obviously give a larger value than multiplying with smaller numbers the same number of times. Thus $7!^{7!}$ is greater than $(7!)!$ and therefore, $j=7$ will give us a fraction as denominator becomes larger than the numerator.

THUS $j=6$ is the final answer. Thanks.