What is the last non-zero digit of $(\dots((2018\underset{! \text{ occurs }1009\text{ times}}{\underbrace{!)!)!\dots)!}}$?

Question

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!

Answer 1

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.