Prove that $(nk)!$ is divisible by $(n!)^k \cdot k!$

Question

I want to prove that $\forall n, k \in \Bbb{N}^*$ $(nk)!$ is divisible by $(n!)^k \cdot k!$

Any ideas?

My attempts:

1.

I failed with attempt to use Mathematical induction like that (for $n$):

$1$: $(1*k)!$ is obviously divisible by $1!^k \cdot k!$
$n$: $A \cdot (n!)^k \cdot k! = (nk)!$, where $A \in \Bbb{N}$
$n+1$: $$ B \cdot ((n+1)!)^k \cdot k!=((n+1)k)! \text{, where } B \in \Bbb{N} $$ $$ A \cdot ((n+1)!)^k \cdot k! = A \cdot (n!)^k \cdot k! (n+1)^k = (nk)! \cdot (n+1)^k $$ And idk if it is possible to achieve something from that, because of this A and B.

2.

Later I tried to proof it in direct way: $$(n!)^k=\prod_{i \in B}{i^k} \text{ , where } B=[1,n] \cap \Bbb{N}$$ $$(nk)!=\prod_{i \in A}{i} \text{ , where } A=[1,nk] \cap \Bbb{N}$$ And I tried to show that $\forall x \in B$ $\exists A_x \subset A$ with $k$ divisible by $x$ elements, e.g. $A_x=\{i \cdot x: i \in [1,k]\}$, but for proof to be right this subsets should be disjoint: $\bigcap_{x \in B} A_x=0$ but I can't invent such subsets :(. Also tried to divide A with remainders, but I didn't get anything sensible

Answer 1

By induction on $k$ (not on $n$), for $n\in\mathbb N$:

For $k=1$, $(n\cdot1)!$ is obviously divisible by $(n!)^1\cdot 1!$.

Assume that $(nk)! = A(n!)^k\cdot k!$ for some $k\in\mathbb N$ and $A\in\mathbb Z$.

Then for $k+1$,

$$\begin{align*} [n(k+1)]! &= (nk)!\cdot(nk+1)(nk+2)\cdots(nk+n-1)(nk+n)\\ &= \left[A(n!)^k\cdot k!\right] \cdot (nk+1)(nk+2)\cdots(nk+n-1)n(k+1)\\ &= A(n!)^k \cdot \underbrace{(nk+1)(nk+2)\cdots(nk+n-1)}_{(n-1)\text{ consecutive integers, multiple of }(n-1)!}n \cdot (k+1)!\\ &= A'(n!)^{k+1}\cdot (k+1)! \end{align*}$$

Using the property that the product of $n-1$ consecutive integers is divisible by $(n-1)!$. So the given divisibility statement doesn't hold for $n=0$.