If $\frac{99 !}{101 !-99 !}=\frac{1}{n}$, determine the value of $n$.

(If $m$ is a positive integer, then $m$ ! represents the product of the integers from 1 to $m$, inclusive. For example, $5 !=5(4)(3)(2)(1)=120$ and $99 !=99(98)(97) \cdots(3)(2)(1)$.
Note that $99 !=99(98)(97) \cdots(3)(2)(1)$ and $101 !=101(100)(99)(98)(97) \cdots(3)(2)(1)$.

Thus, $101 !=101(100)(99 !)$.

Therefore,

$$
\frac{99 !}{101 !-99 !}=\frac{99 !}{101(100)(99 !)-99 !}=\frac{99 !}{99 !(101(100)-1)}=\frac{1}{101(100)-1}=\frac{1}{10099}
$$

Therefore, $n=10099$.

ANSWER: $n=10099$