If $n$ is a positive integer, the symbol $n$ ! (read " $n$ factorial") represents the product of the integers from 1 to $n$. For example, $4 !=(1)(2)(3)(4)$ or $4 !=24$. Determine

$$
\frac{1}{\log _{2} 100 !}+\frac{1}{\log _{3} 100 !}+\frac{1}{\log _{4} 100 !}+\cdots+\frac{1}{\log _{99} 100 !}+\frac{1}{\log _{100} 100 !}
$$
By the change of base formula for logarithms, if $x$ and $y$ are positive real numbers, then

$$
\frac{1}{\log _{x} y}=\frac{1}{\frac{\log _{10} y}{\log _{10} x}}=\frac{\log _{10} x}{\log _{10} y}=\log _{y} x
$$

By this and the identity $\log x+\log y=\log x y$, we have

$$
\begin{aligned}
\frac{1}{\log _{2} 100 !}+\frac{1}{\log _{3} 100 !}+\cdots+\frac{1}{\log _{100} 100 !} & =\log _{100 !} 2+\log _{100 !} 3+\cdots+\log _{100 !} 99+\log _{100 !} 100 \\
& =\log _{100 !}(2 \times 3 \times \cdots \times 99 \times 100) \\
& =\log _{100 !}(1 \times 2 \times 3 \times \cdots \times 99 \times 100) \\
& =\log _{100 !}(100 !) \\
& =1
\end{aligned}
$$

ANSWER: 1