Prove that $e^n\bmod 1$ is dense in $[0,1]$

Question

I just noticed that I have left unanswered one part of an old multi-part question and so decided to re-ask it separately:

Consider the sequence $e^n\bmod 1$, $n\in\Bbb N$. Show that it is dense in $[0,1]$.

This apparently does require specific (approximation?) properties of $e$, as for example replacing $e$ with any integer leads to a non-dense sequence. On the other hand, for every sequence of numbers $a_n\in(0,1)$, it is not hard to find $\alpha$ such that $|\alpha^{2^n}\bmod 1- a_n|<\frac1n$ for all $n$, or $\beta$ such that $|\beta^n\bmod 1-a_n|<\frac1{1000}$. Hence there exist (irrational) bases that lead to a dense sequence and others that lead to a non-dense sequence. Other than that I'm a bit at a dead end.

Answer 1

I was asked to turn a comment into an answer, even though it was mainly a quote, not any work of my own. It's well known that the fractional parts $\{\theta^n\}$ are not just dense, but uniformly distributed for almost all $\theta$. The irony is, that for any individual $\theta$, we don't know nearly as much. Let's quote http://www.kurims.kyoto-u.ac.jp/~kenkyubu/bessatsu/open/B34/pdf/B34_009.pdf

For instance, we cannot disprove that $\displaystyle \lim\{e^{n}\}=0$,where $\{x\}$ is the fractional part of a real number $x$. In the case where $\alpha$ is a transcendental number, it is generally difficult to prove that the sequence $\{\alpha^{n}\}(n=0,1, \ldots)$ has two distinct limit points.

So there's little hope concerning transcendental numbers like $e$, and the results for algebraic $\theta$ aren't exactly mind-boggling, either. We can but hope that there will be some progress, soon.