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I am conducting a little research on checking if a number, written in positional numeral system is irrational.

Let $h^p_n$ be the most right non-zero digit of number $n!$ written in numeral system with base $p$. For example, in system with base $10$, $8!=40320\Rightarrow h_8^{10}=2.$ The question is following: for which $p$ number $H^p=0,h^p_1h^p_2\ldots h^p_n\ldots$ is irrational?

Obviously, $H^2$ is rational. I also know that $H^{10}$ is irrational, but have no idea on how to prove it and would really appreciate any kind of hint.

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    $\begingroup$ There are recursive formulas for this rightmost nonzero digit. These recursive formulas can probably be used to show that the sequence is not periodic, hence this number will be irrational. $\endgroup$
    – Greg Martin
    Commented Jun 1, 2016 at 17:20
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    $\begingroup$ Have you seen home.wlu.edu/~dresdeng/papers/two.pdf ? $\endgroup$
    – so-called friend Don
    Commented Jun 1, 2016 at 17:23
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    $\begingroup$ If $p$ is prime (so $(p-1)!=-1\pmod{p}$) and $n=\sum_id_ip^i$ in base $p$ then it works out that $h^p_n=\pm\prod_id_i!\pmod{p}$, and this is not periodic even if we ignore the $\pm$ signs. $\endgroup$
    – Neil Strickland
    Commented Jun 1, 2016 at 17:38
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    $\begingroup$ Were it periodic with period $a = p^e b$ with $(b,p)=1$, by multiplying $b$ suitably (since it divides some $p^f - 1$), wlog $b = p^f - 1$, i.e. $a = p^{e+f} - p^e$, i.e. in base $p$ $a$ is a string of $(p-1)$'s followed by a string of $0$'s. Now take $n = 2p^{e+f-1}$, which has base $p$ digits $(0,2,0,\ldots,0,0,\ldots,0)$, and compare $h_n^p$ with the same for $n+a$, which has base $p$ digits $(1,1,p-1,\ldots,p-1,0,\ldots,0)$. One gets that $2! = 2\equiv \pm 1\pmod{p}$, which is false once $p>3$. $\endgroup$
    – alpoge
    Commented Jun 1, 2016 at 20:15
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    $\begingroup$ How, Michael, can you know $H^{10}$ is irrational, if you have no idea how to prove it? $\endgroup$
    – Gerry Myerson
    Commented Jun 1, 2016 at 23:35

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