Open Contests

For $n=3$ the claim holds: $(2n)!=6!=720$ and $n^{2n}=3^6=729$.

Suppose $n\ge 4$. Divide the numbers $2,3,\dots ,2n-2$ into pairs $(k,2n-k)$ with $2\le k\le n-1$, leaving $n$ alone. For each pair we have $$k(2n-k)=(n-(n-k))(n+(n-k))=n^2-(n-k{)}^2<n^2.$$ Hence $2\cdot 3\cdot \cdots \cdot (2n-2)<(n^2{)}^{n-2}\cdot n=n^{2n-3}$, therefore $$(2n)!<1\cdot n^{2n-3}\cdot (2n-1)\cdot (2n)<n^{2n-3}\cdot (2n{)}^2=4n^{2n-1}\le n^{2n}.$$

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Alternative solution.

For $n=3$ the claim holds. Suppose the claim holds for $n$; to show that it also holds for $n+1$ it is enough to show the inequality $(2n+1)(2n+2)<\frac{(n+1{)}^{2n}}{n^{2n}}(n+1{)}^2$.

Since $(2n+1)(2n+2)<(2n+2{)}^2=4(n+1{)}^2$, it is enough to show that $\frac{(n+1{)}^{2n}}{n^{2n}}>4$.

This is equivalent with ${\left(1+\frac{1}{n}\right)}^n>2$ which holds for all $n\ge 2$.