Prove that we can assume $u>1$. The conclusion is equivalent to proving that the equation
$$n!=u^{r}\left(u^{s}-1\right)$$

has at most finitely many positive integer solutions $(n, r, s)$.
First, note that for a given $n$, equation (1) clearly has at most finitely many solutions $(r, s)$. We will prove that when $n$ is sufficiently large, equation (1) has no solutions, which will prove the above conclusion.

Choose a prime $p \nmid u$. We can assume (1) has a solution $n>p$ (otherwise, there is nothing to prove), and let $p^{\alpha} \| n!$, then
$$\alpha=\sum_{l=1}^{\infty}\left[\frac{n}{p^{l}}\right] \geqslant\left[\frac{n}{p}\right]>a n,$$

where $a$ is a (positive) constant that depends only on $p$.
Let the order of $u$ modulo $p$ be $d$ and $p^{k_{0}} \|\left(u^{d}-1\right)$, then by Example 5 of Unit 8, when $\alpha>k_{0}$, the order of $u$ modulo $p^{\alpha}$ is $d p^{\alpha-k_{0}}$. Since $u$ and $p$ are fixed numbers, $k_{0}$ and $d$ are also fixed numbers. If (1) has solutions for sufficiently large $n$, then by (2) we know $\alpha>k_{0}$. From (1) we get
$$u^{s} \equiv 1\left(\bmod p^{\alpha}\right),$$

hence by the property of the order, we have $d p^{\alpha-k_{0}} \mid s$; in particular, $s \geqslant d p^{\alpha-k_{0}}$. Therefore,
$$u^{s}-1 \geqslant u^{d p^{\alpha-k_{0}}}-1>u^{d p^{\alpha-k_{0}}-k_{0}}-1.$$

But when $n$ is sufficiently large, it is easy to see that the right-hand side of the above inequality $\geqslant n^{n}-1$. Hence by (3) we get $u^{s}-1>n!$, and thus $u^{r}\left(u^{s}-1\right)>n!$, so when $n$ is sufficiently large, (1) has no positive integer solutions $(r, s)$. This completes the proof.