Prove that by the Cauchy-Schwarz inequality, we have
$$\sum_{i=1}^{n} \frac{1}{1-\left|x_{i}\right|^{n}} \cdot \sum_{i=1}^{n}\left(1-\left|x_{i}\right|^{n}\right) \geqslant n^{2} \text {. }$$

Therefore, to prove the original inequality, it suffices to prove

That is, to prove $\square \square$
$$\begin{array}{c} \frac{n^{2}}{\sum_{i=1}^{n}\left(1-\left|x_{i}\right|^{n}\right)} \geqslant \frac{n}{1-\prod_{i=1}^{n} x_{i}} \\ n-n \prod_{i=1}^{n} x_{i} \geqslant \sum_{i=1}^{n}\left(1-\left|x_{i}\right|^{n}\right) \\ \sum_{i=1}^{n}\left|x_{i}\right|^{n} \geqslant n \prod_{i=1}^{n} x_{i} \end{array}$$

That is, $\square$
By the AM-GM inequality, the above inequality holds, hence the original proposition is true.