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 $2 \cdot 3 \cdot \dots \cdot (2n-2) < (n^2)^{n-2} \cdot n = n^{2n-3}$, therefore
$$(2n)! < 1 \cdot n^{2n-3} \cdot (2n-1) \cdot (2n) < n^{2n-3} \cdot (2n)^2 = 4n^{2n-1} \le n^{2n}.$$

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 $n \ge 2$.