<s>[INST] Prove $2^{3n}3^n | (4n)!$ Algebraically 
Prove that $2^{3n}3^n | (4n)!$

I can make a combinational prove.
Consider the set $a_1a_1a_1a_1a_2a_2a_2a_2\cdots a_na_na_na_n$ this has permutations equal to 
$$\frac{(4n)!}{4!^n} \\ = \frac{(4n)!}{(2^33)^n} = \frac{(4n)!}{2^{3n}3^n}$$
Since this is a integer number thus  $2^{3n}3^n | (4n)!$   
But I am looking for a Algebraic prove, how to Algebraically Prove it?  
 [/INST] In the integers from $1$ to $4n$, there will be $2n$ even numbers, of which half ($n$) will be divisible by $4$. 
Also in those integers, there will be at least $n$ numbers divisible by $3$.
Since $(4n)!$ multiplies all such integers together, we can see that $2^{3n}$ divides $(4n)!$, as does $3^n$. And as these two numbers are coprime, we can conclude that their product $2^{3n}3^n$ also divides $(4n)!$
By induction:
Base case: $n=1, (4n)! = 4! = 24$ and $2^3\cdot 3^1 = 24 \mid 24$ as required
Hypothesis $2^{3k}\cdot 3^k \mid (4k)!$  
Note that among any $4$ consecutive numbers we must have a multiple of $3$, a multiple of $4$ and a separate multiple of $2$
$\implies 2^3\cdot 3 \mid \big[(4k+1)\cdot (4k+2)\cdot (4k+1)\cdot (4k+4)\big]$
and since  $(4(k+1))! = (4k)!\cdot (4k+1)\cdot (4k+2)\cdot (4k+1)\cdot (4k+4)$
and  $2^{3k}\cdot 3^k \cdot 2^3\cdot 3 \mid (4k)!\cdot (4k+1)\cdot (4k+2)\cdot (4k+1)\cdot (4k+4)$
$\implies 2^{3(k+1)}\cdot 3^{k+1} \mid (4(k+1))! $
as required.
</s>