<s>[INST] 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 ?
 [/INST] Hint
Induction for the step $n+1$:
$$(n+2)^{n}(n+3)^{n+1}=\frac{(n+3)^{n+1}}{(n+1)^{n-1}}(n+1)^{n-1}(n+2)^{n}>3^n(n!)^2\frac{(n+3)^{n+1}}{(n+1)^{n-1}}$$
We may expect
$$\frac{(n+3)^{n+1}}{(n+1)^{n-1}}>3(n+1)^2 \quad (1)$$
in order to finish the induction.
Backing to $(1)$ we have an equivalent expression:
$$\left(\frac{n+3}{n+1}\right)^{n+1}>3 \Leftrightarrow \left(1+\frac{2}{n+1}\right)^{n+1}>3$$
Using the Bernoulli inequality $(1+x)^m \ge1+mx$ for $x>-1$.
Taking $m=n+1$ and $x=\frac{2}{n+1}$ we get what we want.
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