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.