Combinatorial proof that $\frac{({10!})!}{{10!}^{9!}}$ is an integer [duplicate]

I need help to prove that the quantity of this division :

$\dfrac{({10!})!}{{10!}^{9!}}$

is an integer number, using combinatorial proof

You have $10!$ students. You want to divide them into groups of $10$ and line of the groups; in how many different ways can you do this?

You can line up the students in $(10!)!$ different orders. Now imagine that you count them off by tens and mark a line on the ground between groups of $10$; you have $\frac{10!}{10}=9!$ groups of $10$. Since all you care about is the order of the groups, you can allow the students in each group of $10$ to rearrange themselves within the group as they please, and you’ll still have the same lineup of groups. Each group can rearrange itself in $10!$ different orders, and there are $9!$ groups, so there are $10!^{9!}$ different lineups of students that produce the same lineup of groups. The number of lineups of groups is therefore $\frac{(10!)!}{10!^{9!}}$, which of course must be an integer.

You can replace $10$ by $n$ and $9$ by $n-1$ and repeat the argument to show that $\dfrac{(n!)!}{n!^{(n-1)!}}$ is an integer.

Here's a simple non-combinatorial proof. Still eager to see a truly combinatorial one. The denominator contains only powers of $2$, $3$, $5$, and $7$: $$ (10!)^{9!}=(10\cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2)^{9!}=\left(7\cdot 5^2 \cdot 3^4 \cdot 2^7\right)^{9!}=7^{9!}5^{2\cdot 9!}3^{4\cdot 9!}2^{8\cdot 9!}. $$ The numerator, on the other hand, can be written as a product of $10!$ factors, of which $1/7$ are divisible by $7$, $1/5$ are divisible by $5$ (and $1/25$ by $25$), $1/3$ are divisible by $3$ (and $1/9$ by $9$), and $1/2$ are divisible by $2$ (and $1/4$ by $4$, and $1/8$ by $8$). So the numerator is divisible by at least $$10!/7 = (10/7) \cdot 9! > 9!$$ factors of $7$, at least $$10!/5 + 10!/25 = (10/5 + 10/25)\cdot 9! > 2\cdot 9!$$ factors of $5$, at least $$10!/3 + 10!/9 = (10/3 + 10/9)\cdot 9! > 4\cdot 9!$$ factors of $3$, and at least $$10!/2 + 10!/4 + 10!/8 = (10/2 + 10/4 + 10/8)\cdot 9! > 8\cdot 9!$$ factors of $2$. Hence the ratio is an integer.

Per the higher-ranked answer, note that generally ${a!}/{(b!)^{a/b}}$, where $b$ divides $a$, is the number of ways to partition $a$ items into groups of size $b$: $$ N_{a,b}={{a}\choose{b}}\cdot{{a-b}\choose{b}}\cdots{{2b}\choose{b}}\cdot{{b}\choose{b}}=\frac{a!}{(a-b)!b!}\frac{(a-b)!}{(a-2b)!b!}\cdots\frac{(2b!)}{b!}\frac{b!}{b!}=\frac{a!}{(b!)^{a/b}}. $$ This is clearly an integer; and here $a=10!$ and $b=10$.