Prove that for any natural number $n>1$, we have $(2n)!<(n(n+1))^n$ Prove that for any natural number $n>1$, we have $(2n)!<(n(n+1))^n$  
I tried using induction, but I failed in that approach, I rather found it was untrue, but on several case testing, I found it to be true. I am out of ideas here, please help. Thank you.
 A: Note that 
$$(n+  1 + a)(n-a) = n(n+1) + an - a(n+1) - a^2   \le n(n+1)$$
for $a=0, 1, \dots , n-1$ and the inequality is strict for $a\neq 0$.
Now, regroup $(2n)!$ and you are done quickly.
A: Using Bernoulli's Inequality,
$$
\begin{align}
\left(\frac n{n+x}\frac{n+x+1}{n+1}\right)^{n+1}
&=\left(1-\frac x{(n+1)(n+x)}\right)^{n+1}\\
&\ge1-\frac x{n+x}\\
&=\frac n{n+x}\tag{1}
\end{align}
$$
Multiplying both sides of $(1)$ by $\left(1+\frac xn\right)^{n+1}$ yields
$$
\left(1+\frac x{n+1}\right)^{n+1}\ge\left(1+\frac xn\right)^n\tag{2}
$$
For $n\ge3$, $(2)$ implies
$$
\begin{align}
\frac{(n(n+1))^n}{((n-1)n)^{n-1}}
&=n(n+1)\left(\frac{n+1}{n-1}\right)^{n-1}\\
&=n(n+1)\left(1+\frac2{n-1}\right)^{n-1}\\[4pt]
&\ge 4n(n+1)\\[12pt]
&\gt2n(2n-1)\\[4pt]
&=\frac{(2n)!}{(2n-2)!}\tag{3}
\end{align}
$$
Since the inequality is true for $n=2$, induction with $(3)$ gives that
$$
(2n)!<(n(n+1))^n\tag{4}
$$
Furthermore, $(3)$ shows that the ratio of the sides increases by over $1+\frac3{2n}$, and since $\prod\limits_{n=1}^\infty\left(1+\frac3{2n}\right)$ diverges, the ratio of the sides in $(4)$ must grow without bound.
