Proof check: $(4n)!$ is divisible by $2^{3n}3^{n}$ Question: Show that $(4n)!$ is a multiple of $2^{3n}3^{n}$ for all $n$.
Proof: It's easy (involves kinda messy calculation tho) to show by induction that $(4n)!$ is a multiple of $2^{3n}$. Now, since every third number in the sequence $1,2,...,4n$ is divisible by $3$, there are at least $\dfrac{4n}{3}>n$ many 3's. This proves that $(4n)!$ is also a multiple of $3^n$ and the claim follows.
Is this proof valid?
 A: Your proof is fine.
An alternative for the factors of $3$:  we show that one of $m,m+1,m+2$ is divisible by $3$ for any integer $m$, and then we can use that instead for your induction step for powers of $3$.
Basically, you need in your induction step to show that $(4n+1)(4n+2)(4n+3)(4n+4)$ is divisible by both $8=2^3$ and $3$. It is divisible by $8$ obviously - just factor it as:
$$8(4n+1)(2n+1)(4n+3)(n+1)$$
The fact that it is divisible by $3$ follows from my statement above - one of $4n+1,4n+2,4n+3$ is divisible by $3$.
Or you could argue that $$\binom{4n+4}{4}=\frac{(4n+4)(4n+3)(4n+2)(4n+1)}{4\cdot 3\cdot 2\cdot 1}$$ is always an integer, so the numerator is divisible by the denominator. (That is likely the combinatorial argument you referenced in comments - the value:
$$\frac{(4n)!}{24^n}$$
can be considered as a count of the number of ordered partitions of the set $\{1,2,\dots,4n\}$ into $n$ sets of size $4$.)
A: It works. As an alternative, through a well-known formula:
$$\nu_2((4n)!)=\sum_{m\geq 1}\left\lfloor\frac{4n}{2^m}\right\rfloor=\color{red}{3n}+\sum_{m\geq 3}\left\lfloor\frac{4n}{2^m}\right\rfloor\geq 4n-1-\log_2(n) $$
and:
$$\nu_3((4n)!)=\sum_{m\geq 1}\left\lfloor\frac{4n}{3^m}\right\rfloor\geq\color{red}{n}+\sum_{m\geq 2}\left\lfloor\frac{4n}{3^m}\right\rfloor\geq 2n-2-\log_3(n). $$
A: Combinatorial Proof
Let $N$ be the number of ways to select $n$ groups of four from $4n$ people.  Then, if we label the groups by numbers $1$, $2$, $\ldots$, $n$, we can do it in $n!\cdot N$ ways.  Alternatively, for $k=1,2,\ldots,n$, there are $4n-4(k-1)$ people from which four shall be chosen for the group with label $k$, and this can be done in $\binom{4n-4(k-1)}{4}$ ways.  Therefore,
$$n!\cdot N=\prod_{k=1}^n\,\binom{4n-4(k-1)}{4}=\frac{(4n)!}{(4!)^n}\,.$$
Consequently,
$$N=\frac{(4n)!}{(4!)^nn!}$$
is an integer.  Hence,
$$\frac{(4n)!}{(4!)^n}=\frac{(4n)!}{2^{3n}3^n}$$
is an integer, which is divisible by $n!$.
A: Hint: Use the Legendre theorem:
For $n!=p_1^{\alpha_1} p_2^{\alpha_2} \cdots $   then
$$
\alpha_i=\sum_{k=0}^{\infty} \left[ \frac{n}{p_i^k}\right]
$$
A: First, show that this is true for $n=3$:
$(4\cdot3)!=3\cdot11550\cdot(2^{3\cdot3}\cdot3^{3\cdot1})$
Second, assume that this is true for $n$:
$(4n)!=3\cdot{k}\cdot(2^{3n}\cdot3^{n})$
Third, prove that this is true for $n+1$:
$(4n+4)!=$
$\color\red{(4n)!}\cdot(4n+1)\cdot(4n+2)\cdot(4n+3)\cdot(4n+4)=$
$\color\red{3\cdot{k}\cdot(2^{3n}\cdot3^{n})}\cdot(4n+1)\cdot(4n+2)\cdot(4n+3)\cdot(4n+4)=$
$3\cdot{k}\cdot(2^{3n}\cdot3^{n})\cdot8\cdot(4n+3)\cdot(4n+1)\cdot(2n+1)\cdot(n+1)=$
$k\cdot(2^{3(n+1)}\cdot3^{n+1})\cdot(4n+3)\cdot(4n+1)\cdot(2n+1)\cdot(n+1)$

Please note that the assumption is used only in the part marked red.
