let $m$ be a natural number, does always exist a $N\in \mathbb{N}$ such that $m$ or more "$0$" digits (excluding the terminal ones) appears amongs the decimal digits of $n!$ if $n\ge N$?

$\begingroup$ It may be possible to show that for every m, the set of n which have fewer than m occurrences of a particular digit among, say, the leading 75% digits of n! has density 0, but I doubt that the same set is finite, much less whether it can be proved finite. Gerhard "Ask Me About System Design" Paseman, 2014.03.14 $\endgroup$ – Gerhard Paseman Mar 14 '14 at 18:17

$\begingroup$ @Gerhard Why do you doubt it's finite? $\endgroup$ – user25199 Mar 14 '14 at 18:37

$\begingroup$ For the same reason that if you pick a random integer, you can find a prime that does not divide that integer, except in one case. Even though it is highly unlikely that there are infinitely many factorials with fewer than 23 8 digits in their base 10 expansion, it is possible, and my intuition does not rule out the possibility. Gerhard "Why Not Believe The Unlikely?" Paseman, 2014.03.14 $\endgroup$ – Gerhard Paseman Mar 14 '14 at 18:50

2$\begingroup$ If we model the nonterminal digits of $n!$ as a random number of the same size (which should be accurate enough for our purposes), then the probability that none of the digits equals $0$ is like $c^{n\log n}$ for some $c<1$. The sum of this quantity over $n\in\Bbb N$ converges (incredibly fast), and so BorelCantelli suggests that there are only finitely many $n$ for which $n!$ has no nonterminal zeros. A slight elaboration gives the same heuristic that there are only finitely many $n$ for which $n!$ has fewer than $m$ nonterminal zeros, for any $m\in\Bbb N$. $\endgroup$ – Greg Martin Mar 14 '14 at 22:43
Your question is partly answered in this question. All possible leading sequences of digits appear in $n!$, including those with $m$ zeros, so there exist infinitely many $n$. But I think proving it for all $n>N$ for some $N$ is beyond reach of current techniques.