Take a prime $p>10^{100}$ and let $n=2018p$. Clearly $\nu_p(n!)=2018$ so this works.

MarkBcc168 wrote: Take a prime $p>10^{100}$ and let $n=2018p$. Clearly $\nu_p(n!)=2018$ so this works. The restriction simply needs to be $p>2019$, correct?

What notation is $v_p(n!)$?

P adic valuation

The number $n = 2017^2$ works since $2017$ is prime and $v_{2017}(2017^2!) = 2018.$