Every time you hit a power of ten, fracPart(log10(n!)) gets the unit interval finely iterated by n=10^k, 10^k + 1, 10^k + 2, ..., 10^k + sqrt(2 log(10) 10^k).
It's not clear whether there is a number A with decimal expansion (a_1 a_2 ... a_k ) such that there doesn't exist a positive number X for which
X ! = (a_{1} a_{2} ... a_{k} x_{k+1} x_{k+1} ... x_{s} )
I agree though, density in (0,1) of the fracPart(log10(n!)) is a good thing to try to prove, since 10^fracPart(log10(n!)) should have the same digits as 10^intPart(log10(n!))*10^fracPart(log10(n!))=10^log10(n!)=n!
E.g. https://www.wolframalpha.com/input/?i=Table%5B%5Bfrac%28log1...
In general for b << 10^k, we have fracPart[log10((10^k + b)!)] ≈ fracPart[log10((10^k)!) + b(b+1)/(2 log(10) 10^k)]. So it takes sqrt(2 log(10) 10^k) steps for these factorial values to work their way around the unit interval. Increase k and you fill it arbitrarily finely.
Edit: ^^^ To be clear, the approximation there is formed using the linear approximation log(1 + b/10^k) ≈ b/10^k, and basic log multiplication/addition/division formulas.
The spacing is not uniform, it's just bounded by O(1/10^(k/2)). The spacing is, at its greatest, about 1/sqrt(2 log(10) 10^k).
If you knew the number of steps, I think you could only make a lower bound on spacing, not an upper bound. Since at worst all the steps could be clustered really close, except for the last one.
But now I have to think more because it seems like it should actually help the argument that the spacing gets finer and finer as you step, because that is what we want anyway. And is the number of steps actually important to the proof?
The number of steps tells you how big the last step is.