prove that $(n^2)!$ is divisible by $(n!)^{n+1}$, where $n\in \mathbb{N}$ (1) How can we prove that $(kn)!$ is divisible by $(n!)^k$, where $n,k\in \mathbb{N}$
(2) How can we prove that $(n^2)!$ is divisible by $(n!)^{n+1}$, where $n\in \mathbb{N}$
(3) How can we prove that $(k!)!$ is divisible by $(k!)^{(k-1)!}$, where $k\in \mathbb{N}$
$\bf{My\; Try}::(1) $ We know that product of $n$ consecutive integer is divisible by $n!$
So Here  $1.2.3.4........................................n$ is divisible by $n!$
Similarly $(n+1)\cdot(n+2)\cdot(n+3)...........(2n)$ is divisible by $n!$
Similarly $(2n+1)\cdot (2n+2)\cdot ..............(3n)$ is divisible by $n!$
Similarly $(3n+1)\cdot(3n+2)......................(4n)$ is divisible by $n!$
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Similarly $\{(k-1)n+1\}\cdot \{(k-1)n+2\}\cdot..................\{(k-1)n+n\}$ is divisible by $n!$
So we can say that $(kn)!$ is divisible by $(n!)^k$
Now I did not understand how can i  solve $(II)$ one and $(III)$ one
Help Required
Thanks
 A: $$ \frac{(n^2)!}{n!^{n+1}}$$
is the number of paritions of an $n^2$ element set into $n$ (indistinguishable) sets of size $n$ each.
$(3)$ is just a special case of $(1)$ with $k\leftarrow (k-1)!$ and $n\leftarrow k$.
A: (3) is a direct consequence of (1), as (after a change of name to avoid confusion) $(h!)! = (h(h-1)!)!$ so applying (1) with $n=h$ and $k=(h-1)!$, you get that $(h(h-1)!)!$ is divisible by $(h!)^{(h-1)!}$.
For (2), you have actually that
$$(k+1)n! \mid (kn+1)⋅(kn+2)⋅\dots·((k+1)n)$$
is divisible by $(k+1)n!$, in fact, if you take out the last term, you get
$$(n-1)! \mid (kn+1)⋅(kn+2)⋅\dots·((k+1)n-1)$$
just because it's the product of $(n-1)$ consecutive integers. Then you multiply by $(k+1)n$ both sides, obtaining that 
$$(n-1)!·(k+1)n \mid (kn+1)⋅(kn+2)⋅\dots·((k+1)n-1)·(k+1)n$$
So $n^2!$ can be expressed as a product of
$$
\prod_{k=0}^{n-1}\prod_{i=1}^{n}kn+i = \prod_{k=0}^{n-1}P_k
$$
And each $P_k$ is divisible by $(k+1)n!$ so the entire product is divisible by
$$
\prod_{k=0}^{n-1}(k+1)n! = n!^{n+1}
$$
A: Hint 1: Consider the multinomial coefficient
$$kn \choose n, n,\dots,n$$
that is an integer.
Hint 2: Follow the @Hagen von Eitzen's method!
Hint 3: Consider the multinomial coefficient
$$k! \choose k,k,\dots,k$$
A: 3 is simple,
plug n = k, m = (k - 1)! into $(mn)!$  is divisible by $(n!)^m$
Going to submit answer to two shortly.
Edit
I am trying to use a proof by induction for 2.
The 0th case (1) is simple $(1^2)! = 1$ and $(1!)^{1+1} = 1$ obviously  $1 | 1$ so that case works. 
All that needs to be done is to prove that $(n^2)! | (n!)^{n+1} \rightarrow ((n+1)^2)! | ((n+1)!)^{n+2}$
