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?

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.
