What is the last non-zero digit of $(\dots((2018\underset{! \text{ occurs }1009\text{ times}}{\underbrace{!)!)!\dots)!}}$?
First, note the following: Let $N$ be an multiple of 10 i.e., $N=10k$. Then the last nonzero digit of $N!$ is
a) $3 \cdot 4 \cdot 6 \cdot 7 \cdot 8 \cdot 9 \equiv _{10} $ 8 if $k \equiv_4 1$;
b) $8^2$ mod 10 which is 4 if $k \equiv_4 2$;
c) $8^3$ mod 10 which is 2 if $k \equiv_4 3$; and
d) $8^4$ mod 10 which is 6 if $k \equiv_4 0$.
Clearly $(\ldots ((2018!)!)! \ldots ) !$ is of the form $10k$; $k$ a multiple of 4. So the last nonzero digit is 6.