Without using computer programs, can we find the last non-zero digit of

$$(\dots((2018\underset{! \text{ appears }1009\text{ times}}{\underbrace{!)!)!\dots)!}}?$$

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 $!$ appears $1009$ 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!

  • 1
    $\begingroup$ have you tried smaller repeats ? $\endgroup$
    – user645636
    Commented Aug 26, 2019 at 14:45
  • 1
    $\begingroup$ smaller inputs @Gabe . With start value 3 I can get $((3!)!)!$ but the next gives me a truncation error in PARI/GP. $\endgroup$
    – user645636
    Commented Aug 26, 2019 at 15:06
  • 5
    $\begingroup$ Probably useful: math.stackexchange.com/questions/130352/… $\endgroup$ Commented Aug 26, 2019 at 15:18
  • 2
    $\begingroup$ @Hussain-Alqatari would you tell me about the source of the problem? $\endgroup$
    – Vulch
    Commented Mar 24 at 8:12
  • 1
    $\begingroup$ @Sbsty It can indicate how hard the problem is or which methods are needed to solve it. For example, it could appear in a computer science / algorithms textbook, or a number theory textbook, or an "open problems" section in a mathematical journal, or maybe OP's friend asked the problem and does not know how hard it is. $\endgroup$ Commented Mar 30 at 14:47


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