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?](https://mathoverflow.net/questions/296321/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.



Is the sequence $(\log(n!)\mod1)_{n\in\mathbb N}$ dense in the interval $[0,1]$?
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.


