13. $[\mathbf{9}]$ Find the smallest positive integer $n$ for which
$$
1!2!\cdots(n-1)!>n!^{2} .
$$
Answer: $\square$
Dividing both sides by $n!^{2}$, we obtain
$$
\begin{aligned}
\frac{1!2!\ldots(n-3)!(n-2)!(n-1)!}{[n(n-1)!][n(n-1)(n-2)!]} & >1 \\
\frac{1!2!\ldots(n-3)!}{n^{2}(n-1)} & >1 \\
1!2!\ldots(n-3)! & >n^{2}(n-1)
\end{aligned}
$$

Factorials are small at first, so we can rule out some small cases: when $n=6$, the left hand side is $1!2!3!=12$, which is much smaller than $6^{2} \cdot 5$. (Similar calculations show that $n=1$ through $n=5$ do not work. either.)
Setting $n=7$, the left-hand side is 288 , which is still smaller than $7^{2} \cdot 6$. However, $n=8$ gives $34560>448$, so 8 is the smallest integer for which the inequality holds.