How Do I Simplify These Fractions?

We will look at the ratio of these fractions with each other: $\begin{aligned} \dfrac{\frac{100!}{101}}{\frac{101!}{102}}&= \dfrac{102}{101^2}<1\implies \boxed{\dfrac{101!}{102}>\dfrac{100!}{101}}\\ \dfrac{\frac{101!}{102}}{\frac{102!}{103}}&= \dfrac{103}{102^2}<1\implies \boxed{\dfrac{102!}{103}>\dfrac{101!}{102}}\\ \dfrac{\frac{102!}{103}}{\frac{103!}{104}}&= \dfrac{104}{103^2}<1\implies \boxed{\dfrac{103!}{104}>\dfrac{102!}{103}}\end{aligned}$ Now we have the following chain inequality: $\dfrac{103!}{104}>\dfrac{102!}{103}>\dfrac{101!}{102}>\dfrac{100!}{101}$ Therefore the largest number is $\boxed{\dfrac{103!}{104}}$