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.

$x_n \operatorname{mod1} =\big{ x_n \big}$ where $\big{ x_n \big}= x_n -[x_n].$ — Rick Sanchez C-666, Nov 18 '20 at 17:38
"For the exponential this is a bit more difficult." In fact, it's a bit too difficult for mathematics as we know it, today. — , Nov 18 '20 at 17:55
@ProfessorVector :) Some references for an "ignoramus" would also make an appreciated asnwer — Hagen von Eitzen, Nov 18 '20 at 20:02
@ProfessorVector Your comment was directed toward the one above yours, right? You are not saying that the problem can't be solved. — John Douma, Nov 18 '20 at 20:38
http://www.kurims.kyoto-u.ac.jp/~kenkyubu/bessatsu/open/B34/pdf/B34_009.pdf — , Nov 18 '20 at 22:11
@ProfessorVector Thanks! I'd still suggest to turn this into an answer so that this question finds some kind of acceptable completion. That won't prevent future users from adding something definite in a few years :) — Hagen von Eitzen, Nov 19 '20 at 21:08
@JohnDouma The comment was directed my answer to the linked multipart question and kindly shows that my light-hearted and confident handling of a gap was rather an understatement — Hagen von Eitzen, Nov 19 '20 at 21:12

score 6 · Accepted Answer · answered Nov 20 '20 at 11:47

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.

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

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