We have $n^{3}-1=(n-1)\left(n^{2}+n+1\right)$. We shall prove that there are infinitely many positive integers $n$ such that $n^{2}+n+1$ can be represented as the product of some factors less than $n-1$ and then deduce that $n^{3}-1$ divides $n!$. In fact, we first choose $n=m^{2}$ for some positive integer $m$, one has
$$n^{2}+n+1=m^{4}+m^{2}+1=\left(m^{2}+1\right)^{2}-m^{2}=\left(m^{2}-m+1\right)\left(m^{2}+m+1\right) .$$
Next, we choose $m=3k+1$, then
$m^{2}+m+1=(3k+1)^{2}+(3k+1)+1=9k^{2}+6k+1+3k+1+1=3\left(3k^{2}+3k+1\right)$
and
$m^{2}-m+1=(3k+1)^{2}-(3k+1)+1=9k^{2}+6k+1-3k-1+1=9k^{2}+3k+1$.
It is clear that
$$3<3k^{2}+3k+1<9k^{2}+3k+1<9k^{2}+6k=(3k+1)^{2}-1=n-1 .$$
Therefore, for $n=(3k+1)^{2}$ with any positive integer $k$, it follows that $n!$ is divisible by
$$3\left(3k^{2}+3k+1\right)\left(9k^{2}+3k+1\right)\left(9k^{2}+6k\right)=n^{3}-1 .$$
The problem is proved. $\square$