<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] prove: $(n+1)^{n−1}(n+2)^n>3^n(n!)^2=3^n(n!n!)$ for $n>1$
$n=2:3^14^2=48>3^2(2)(2)=36$
assume: $(n+1)^{n−1}(n+2)^n>3^n(n!n!)$
need to arrive at: $(n+2)^{n}(n+3)^{n+1}>3^{n+1}(n+1)!(n+1)!$
for lhs need to multiply by: 
${{(n+2)^{n}(n+3)^{n+1}}\over {(n+1)^{n−1}(n+2)^n}}
={{(n+3)^n(n+3)(n+1)}\over {(n+1)^n}}=({{n+3}\over {n+1}})^n(n+3)(n+1)$
where $({{n+3}\over {n+1}})^n$ is greater than $3$ for $n=3$ and increasing for all $n\in N$. 
for rhs need to multiply by: $3(n+1)(n+1)$.  therefore lhs > rhs and 
$(n+1)^{n−1}(n+2)^n>3^n(n!)^2$ for $n>1$ by mathematical induction.
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