7. Let $m$ and $n$ both be positive integers greater than 1. Given that $\frac{(n+m)!}{n!}=17297280$. Then $\frac{n}{m}=$ $\qquad$
7.1 or $\frac{31}{2}$.

Since $\frac{(n+m)!}{n!}=17297280$, factorizing it we get
$$
\begin{array}{l}
17297280=10 \times 1001 \times 27 \times 64 \\
=2^{7} \times 3^{3} \times 5 \times 7 \times 11 \times 13, \\
\frac{(n+m)!}{n!}=(n+1)(n+2) \cdots(n+m),
\end{array}
$$

Therefore, it can be decomposed into two cases, i.e.,
$$
\begin{array}{l}
(n+1)(n+2) \cdots(n+m) \\
=8 \times 9 \times 10 \times 11 \times 12 \times 13 \times 14,
\end{array}
$$

or $(n+1)(n+2) \cdots(n+m)$
$$
=63 \times 64 \times 65 \times 66 \text {. }
$$

Thus, $\left\{\begin{array}{l}n+1=8, \\ n+m=14\end{array} \Rightarrow \frac{n}{m}=1\right.$,
or $\left\{\begin{array}{l}n+1=63, \\ n+m=66\end{array} \Rightarrow \frac{n}{m}=\frac{31}{2}\right.$.