9. If positive integers $m, n$ satisfy $\frac{(m+n)!}{n!}=5040$, then the value of $m!n$ is $\qquad$

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9. 144 .
$$
\begin{array}{l}
\text { Given } \frac{(m+n)!}{n!} \\
=(m+n)(m+n-1) \cdots(n+1) \text {, } \\
5040=10 \times 9 \times 8 \times 7,
\end{array}
$$

we know $\left\{\begin{array}{l}m+n=10, \\ n+1=7\end{array} \Rightarrow\left\{\begin{array}{l}m=4 \text {, } \\ n=6 .\end{array}\right.\right.$
Thus, $m!n=144$.