Why is $(n+1)^{n-1}(n+2)^n>3^n(n!)^2$ 
Why is $(n+1)^{n-1}(n+2)^n>3^n(n!)^2$ for $n>1$

I can use $$(n+1)^n>(2n)!!=n!2^n$$ but in the my case, the exponent is always decreased by $1$, for the moment I don't care about it, I apply the same for $n+2$
$(n+2)^{n+1}>(2n+2)!!=(n+1)!2^{n+1}$  
gathering everything together,
$(n+1)^{n-1}(n+2)^n=\frac{(n+1)^n(n+2)^{n+1}}{(n+1)(n+2)}>\frac{(n+1)(n!)^22^{2n+1}}{(n+1)(n+2)}$
$\iff(n+1)^{n-1}(n+2)^n>(n!)^2\times\frac{2^{2n+1}}{(n+2)}$
but $\frac{2^{2n+1}}{(n+2)}>3^n$ is not true for $n=2$
can you suggest another approach ?

EDIT: This answer is wrong, because I mixed up my left and right-hand sides right at the end. I think it is salvageable, but it'll be quite a bit of work.
I'll do it without induction.
Rearrange: we want $\left(\frac{(n+1)(n+2)}{3}\right)^{n-1} \frac{n+2}{3} > (n!)^2$
We'll show that this actually holds if we remove the $\frac{n+2}{3}$ term (which is always $\geq 1$ anyway).
The right-hand side of the modified desired inequality is is $$2^2 \times 3^2 \times \dots \times n^2$$ with $n-1$ terms.
The left-hand side is $$\frac{(n+1)(n+2)}{3} \times \dots \times \frac{(n+1)(n+2)}{3}$$ with $n-1$ terms again.
But $\frac{(n+1)(n+2)}{3}$ is bigger than or equal to the $i^2$ term whenever $i$ is less than or equal to $\sqrt{\frac{(n+1)(n+2)}{3}}$, and that's $\leq n$ whenever $n > 2$. So if $n>2$, for every $i \leq n$ we have each right-hand term less than its corresponding left-hand term.
So we only need to check $n=1$ and $n=2$, and they're very easy.
