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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?

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    Have you thought of trying induction? Seems like a straightforward induction proof to me – 2017-01-18
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    What @JMoravitz said. But in plain English, if you increase $n$ by $1$, you multiply the left side by $24$ and you multiply the right side by at least one multiple of $3$, exactly one multiple of $4$, and exactly one other multiple of $2$. – 2017-01-18
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    or, one can directly use the fact that the product of $k$ consecutive numbers is divisible by $k!$ – 2017-01-18
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    The tag ([tag:theorem-provers]) is for questions about software designed for checking formal proofs or assisting with writing them, see the [tag-wiki](http://math.stackexchange.com/tags/theorem-provers/info). It is not intended for all questions which are about proofs of theorems. – 2017-01-19

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Hint: Among four consecutive numbers each of $0,1,2,$ and $3\pmod{4}$ will occur.

Among four consecutive numbers at least one will be divisible by $3$.

Finally, note that $(4(n+1))!=(4n)!(4n+1)(4n+2)(4n+3)(4n+4)$

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    All of your solutions are nice :D Just found this can be easily proved by induction and also "the product of k consecutive numbers is divisible by k!".. Thanks :) – 2017-01-18
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you have to prove $v_2((4n)!)\geq 3n$ and $v_3((4n)!)\geq n$.

This is easy if you use Polignac's formula:

$V_p(x!)=\sum\limits_{n=1}^\infty \lfloor x/p^n \rfloor$

We get :

$v_2((4n)!)=\sum\limits_{n=1}^\infty \lfloor 4n/2^n\rfloor \geq 2n+n=3n$.

$v_3((4n)!)=\sum\limits_{n=1}^\infty \lfloor 4n/3^n \rfloor\geq \lfloor 4n/3 \rfloor \geq n $

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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.

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Notice If $3a < 4n$ if $a \le n$ so there are at least $n$ multiples of $3$ that are less than $4n$ so they are terms that go into $n! = 1*2*[3]*....*[3i]*....*[3n]*...... *4n$ so $3^n|4n!$

Likewise for every $1 \le a\le n$ you have the $n$ pairs of terms $4a-2$ and $4a$ so $2^n4^n = 2^{3n}|4n!$.

So $\text{least common multiple}(3^n, 2^{3n}) = 2^{3n}3^n|4n!$.