The claim is equivalent to $$(2n+1)^n - (2n-1)^2 \ge (2n)^n$$

This relation indeed holds because $$\begin{align*} & (2n+1)^n - (2n-1)^n \\ =\ &2\left((2n)^{n-1}\binom{n}{1} + (2n)^{n-3}\binom{n}{3} + \cdots\right) \\ \ge\ & 2\cdot (2n)^{n-1}\binom{n}{1}\\=\ &(2n)^n \end{align*}$$