Is the sequence $(\log(n!)\mod1)_{n\in\mathbb N}$ dense in the interval $[0,1]$? This question was raised in the comment by Todd Trimble at how to proof there is a natural number n, the first four digits of n! Is 2018?. I thought the question may be posted separately, as even partial answers to it may turn out useful and instructive for a number of users (certainly including me).  
A trivial remark is that, in view of Stirling's formula, this question can be restated as follows: Is the sequence $((n+1/2)\log n-n\log e\mod1)_{n\in\mathbb N}$ dense in the interval $[0,1]$? It would be surprising to me if the answer here is no or if it depends on the base of the logarithm.
 A: Let $a$ be the base of logarithm here. Consider the function $$f(x)=\frac{(x+1/2)\ln x-x}{\ln a}.$$
As noted in the question, it is enough to prove that the sequence $f(n)$ is dense modulo $1$. In fact, this sequence is equidistributed. By Weyl's criterion, it is enough to show that for any nonzero integer $k$ we have
$$\sum_{n \leq N} e^{2\pi ikf(n)}=o(N).$$
To prove this, let us use the van der Corput lemma, which states that if $g \in C^2(I)$, where $I$ is an interval and
$$\lambda\ll |g''(x)| \ll \lambda$$
for all $x \in I$ (constants in Vinogradov symbols are absolute) then
$$\sum_{n \in I\cap \mathbb Z} e^{2\pi ig(n)} \ll |I|\lambda^{1/2}+\lambda^{-1/2}.$$
Taking $g(x)=kf(x)$ and $I=[M,M_1]$ with $M<M_1 \leq 2M$ we get for $M$ large enough
$$\sum_{M\leq n\leq M_1} e^{2\pi i kf(n)} \ll \sqrt{kM},$$
as $f''(x)=\frac{1}{x\ln a}-\frac{1}{2x^2\ln a} \asymp \frac{1}{M}$ for $x \in I$.
Therefore, using the dyadic subdivision of the interval $[1,N]$ we get
$$\sum_{n \leq N} e^{2\pi i kf(n)} \ll \sqrt{kN}+1=o(N),$$
as needed.
