Without using computer programs, can we find the last non-zero digit of $(\dots((2018\underset{! \text{ occurs }1009\text{ times}}{\underbrace{!)!)!\dots)!}}$?
What I know is that the last non-zero digit of $2018!$ is $4$, but I do not know what to do with that $4$.
Is it useful that $!$ occurs $1009$ times where $1009$ is half of $2018$? If that is useful, then what if $1009$ was another value, say $1234$?
Any help will be appreciated. THANKS!