**Proof that the power of $2$ in $(3n)!$ is greater than or equals to the power of $2$ in $n!(n+1)!(n+2)!$.**

I tried doing some algebraic manipulation,

$\frac{(3n)!}{n!(n+1)!(n+2)!}=\binom{(3n)!}{(n+2)!}\frac{(2n-2)!}{(n+1)!n!}=\binom{(3n)!}{(n+2)!}\binom{(2n-2)!}{(n+1)!}\frac{(n-3)!}{n!}=\binom{(3n)!}{(n+2)!}\binom{(2n-2)!}{(n+1)!}\frac{1}{(n-2)(n-1)n}$

Here, $\binom{(3n)!}{(n+2)!}$ and $\binom{(2n-2)!}{(n+1)!}$ are integers but $\frac{1}{(n-2)(n-1)n}$ is creating problems.

My argument is that some powers of $2$ from $\binom{(3n)!}{(n+2)!}$ and $\binom{(2n-2)!}{(n+1)!}$ would cancel out the powers of $2$ from $\frac{1}{(n-2)(n-1)n}$. *But the argument is too ambiguous to be written down as a "proof".*

Please let me know if there's a better way to approach these kinds of problems.

*Any help would be highly appreciated.*