Nonzero digits in n!

Can it be shown that a positive fraction of the base-$b$ digits of n! are nonzero (in the limit as $n\to\infty$)?

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The answer: $9/10$ of course but even $> 0$ is most possibly unprovable presently. I asked a similar question about powers of an integer (say, $3$) here: mathoverflow.net/questions/38971/… (see Update 4 in that question). – Mark Sapir Dec 24 '11 at 0:10
I meant $(b-1)/b$, of course. – Mark Sapir Dec 25 '11 at 14:08
i.e. $\ 9 := 10-1,\$ whatever $\ 10\$ would be. – Włodzimierz Holsztyński Nov 3 '14 at 5:02

The bound given by F. Luca and cited in Gjergji Zaimi's answer has been recently improved, although only by a factor $\log \log \log n$. Precisely, in C. Sanna, On the sum of digits of the factorial, J. Number Theory 147 (2015), 836--841, the author proved that $$\min\{s_b(n!),s_b(\Lambda_n)\} > c_b \log n \log \log \log n ,$$ for all integers $n > e^e$, where $s_b(\cdot)$ is the sum of base-$b$ digits function, $c_b > 0$ is a constant depending only on $b$, and $\Lambda_n$ is the least common multiple of $1, \ldots, n$. Clearly, this inequality also holds with $s_b$ replaced by $t_b$, i.e., the function counting the number of non-zero base-$b$ digits of its argument, since $t_b(m) \geq s_b(m)/b$ for each positive integer $m$.
I believe the best current lower bound on this is the one given by F. Luca in "The Number of Non-Zero Digits of n!" Canad. Math. Bull. 45(2002), 115-118. It is proven there that the number of non-zero base $b$ digits grows at least as fast as $C_b\log n$.
Interesting! It is better than the current lower bound for $a^n$ (where $a$ is co-prime with 10). – Mark Sapir Dec 24 '11 at 0:39