Proof: Let $p_{1}<p_{2}<\cdots<p_{k}$ be all the prime numbers not exceeding $n$, and let $q$ be the smallest prime number greater than $n$. If $n=2$, then $q=3$, and the proposition holds. If $n=3$, then $q=5$, and the proposition holds. When $n>3$, $2,3$ both appear in $p_{1}, \cdots, p_{k}$, so $5 \leqslant q \leqslant n!-1n$, thus $n<p \leqslant q<n!$. Therefore, the proposition holds.