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Sep 1, 2014 at 9:35 comment added user40023 I have read [1], you remark that Luca noted that his lower bound is probably weaker, but this is actually the point of my question! It is true that many of the less significant digits of $n!$ are zeros, but not too many: by Legendre-De Polignac formula the number $Z_n$ of trailing zeros of $n!$ is less than $n/(p-1)$, where $p$ is the greatest prime factor of $n$, so assuming that $n! / b^{Z_n}$ is "generic" my conjecture gives again $s_b(n!) = s_b(n! / b^{Z_n}) > C n \log n$. Thus I downvote you answer.
Sep 1, 2014 at 6:49 history answered Aaron Meyerowitz CC BY-SA 3.0