Is that too big to calculate?

$\begin{aligned} S & = \frac 1{\log_2 2019!} + \frac 1{\log_3 2019!} + \frac 1{\log_4 2019!} + \cdots + \frac 1{\log_{2019} 2019!} & \small \color{#3D99F6} \text{By }\log_b a = \frac {\log a}{\log b} \\ & = \frac {\log 2}{\log 2019!} + \frac {\log 3}{\log 2019!} + \frac {\log 4}{\log 2019!} + \cdots + \frac {\log 2019}{\log 2019!} \\ & = \frac {\log 2 + \log 3 + \log 4 + \cdots + \log 2019}{\log 2019!} & \small \color{#3D99F6} \text{As }\log a + \log b + \log c + \cdots + \log z = \log (abc\cdots z) \\ & = \frac {\log 2019!}{\log 2019!} = \boxed 1 \end{aligned}$

Each of the terms of the sum can be rewritten using $\frac{1}{\log_b 2019!}=\frac{\log b}{\log 2019!}$

Hence the sum is $\frac{\log2+\log3+\cdots +\log{2019}}{\log 2019!}=\frac{\log 2019!}{\log 2019!}=\boxed1$