Floor of a number difficult to compute

In the inequality $n! \leq e\;n^{n+\frac{1}{2}} e^{-n}$ (see the Stirling's approximation ), make $n=10^{2018},$ and use the fact that $2<e<3.$ Then you get $(10^{2018})!\leq e (10^{2018})^{(10^{2018}+\frac{1}{2})} (e^{-10^{2018}})< 3*(\frac{10^{2018}}{2})^{10^{2018}} 10^{1009}=3(5^{10^{2018}}) 10^{2017*10^{2018}+1009}.$ Dividing both sides by the number in the right side of the inequality, $\frac{(10^{2018})!}{3(5^{10^{2018}}) 10^{2017*10^{2018}+1009}}<1.$ Therefore, $\Bigl\lfloor \frac{(10^{2018})!}{3(5^{10^{2018}}) 10^{2017*10^{2018}+1009}}\Bigr\rfloor=\boxed{0}$