How can we prove that $\displaystyle \frac{n^2!}{(n!)^2}\in \mathbb{Z}\;,$ Where $n\in \mathbb{N}$
$\bf{My\; Solution::}$ Using The formula $n^2! = (2n)!\cdot (2n+1)\cdot(2n+2)\cdot\cdot \cdot \cdot n^2$
So $\displaystyle \frac{n^2!}{(n!)^2} = \frac{(2n)!\cdot (2n+1)\cdot (2n+2)...........(n^2)}{n!\cdot n!}=\binom{2n}{n}\times \bf{Integer\; quantity}\in \mathbb{N}$
My Question is How can we solve combinatorial way or any other way.
Plz explain me
Thanks
Remark : For every (prime) number $p$: $$ n \cdot \lfloor \dfrac{n}{p^{\alpha}} \rfloor \leq \lfloor \dfrac{n^2}{p^{\alpha}} \rfloor \ . $$ Proof : Let's denote by $m$ and $p$ the integral part and fractional part of $\dfrac{n}{p^{\alpha}}$
$$ m=\lfloor \dfrac{n}{p^{\alpha}} \rfloor \ \ \ \ \ \text{and} \ \ \ \ \ p=\lfloor \dfrac{n}{p^{\alpha}} \rfloor \ ; $$ then one can see easilly that : $$ \dfrac{n}{p^{\alpha}}=m+p \Longrightarrow \dfrac{n^2}{p^{\alpha}}=n \cdot m + n \cdot p \Longrightarrow \\ n \cdot \lfloor \dfrac{n}{p^{\alpha}} \rfloor = n \cdot m = \lfloor n \cdot m \rfloor \leq \lfloor \dfrac{n^2}{p^{\alpha}} \rfloor \ . $$
Remark (II) : For every prime number $p$: $$ n \cdot \sum \lfloor \dfrac{n}{p^{\alpha}} \rfloor \leq \sum \lfloor \dfrac{n^2}{p^{\alpha}} \rfloor \ \ \Longrightarrow \\ \ \ \ \ \ n \cdot v_p(n!) \leq v_p\Big((n^2)!\Big) \Longrightarrow \\ \ \ \ v_p(n! ^n) \leq v_p\Big((n^2)!\Big) \ . $$
Remark (III) : $$ n! ^n \mid (n^2)! \ . $$
This is a combinatorial proof.
Suppose you have $n^2$ chairs arranged as a square of $n\times n$, the way $n^2$ persons sit down is $n^2!$.
Choose $n$ person for the first row, and $n$ person for the second row, and arrange their seats, you get ${n^2 \choose n }{n^2 - n \choose n} (n!)^2$.
Since for each fixed arrangement, the way of arranging the other $n^2 - 2n$ persons in the rest $n-2$ rows is the same, we get $\dfrac{n^2!}{{n^2 \choose n }{n^2 - n \choose n} (n!)^2}$ is an integer.
So $\dfrac{n^2!}{(n!)^2} = \dfrac{n^2!}{{n^2 \choose n }{n^2 - n \choose n} (n!)^2}{n^2 \choose n }{n^2 - n \choose n} $ is an integer.
I dont know if this maybe useful for you as a "prove" but any rising or falling factorial of length n is divisible by n! because the length itself determine the periodicity (and order) of the factors.
If you have a list of r consecutive numbers in the list will exist at least one number that is divisible for all numbers from 1 to r. By example in the list 11,12,13,14,15 it will be a number divisible by 2, other by 3, other by 4 and other by 5 because the length of the list is 5 and multiples of any number have it periodicity.
Another example: if I have a list of consecutive numbers of length 13 someone must be divisible by 13, some other by 12, 11, 10, 9... And they have a order so they cant overlap over the same number, so one number will be divisible by 13 and the same time another will be divisible by 12, an a different one by 11 and so on.
You can see that $\frac{n^2!}{(n!)^2}=\frac{1\cdot 2\cdot 3\cdots n^2}{n!\times n!}=\frac{(n^2)_n}{n!}$. The length of the consecutive multipliers on the numerator is the same that the consecutive primary multipliers in the denominator so this fraction must be a natural number.
A more interesting thing to see is what happen for the cases of custom factorials of step h.
