Is $\{ \sin n^m \mid n \in \mathbb{N} \}$ dense in $[-1,1]$ for every natural number $m$? Is $\{\sin n^m \mid n \in \mathbb{N}\}$ dense in $[-1,1]$ for every natural number $m$?

Progress
For $m=1$, I can prove this using the fact that $\sin$ is continuous and $a+b\pi$ is dense in the real line, for integer $a$ and $b$. However, this approach breaks down for $m>1$; what method should be used then?
 A: The answer is yes.
Given any real number $x_0$, the map $T_{x_0} : (x_1,\ldots,x_d) \mapsto (x_1+x_0,x_2+x_1,\ldots,x_d+x_{d-1})$ from $(\mathbf{R}/\mathbf{Z})^d$ to itself preserves the Haar probability measure. When $x_0$ is an irrational number, it is ergodic. This can be checked by using Fourier series.
A recursion shows that for every integer $n \ge 0$ and $(x_1,\ldots,x_d) \in (\mathbf{R}/\mathbf{Z})^d$,
$$T_{x_0} ^n(x_1,\ldots,x_d) = \Big(x_1+x_0~,~x_2+nx_1+{n \choose 2}x_0~,~\ldots~,~\sum_{k=0}^d {n \choose k} x_{d-k}\Big).$$
Applying Birkoff ergodic theorem to any integrable fonction depending on the last variable $x_d$ only yields the following statement.
If $P$ is a non-constant polynomial whose leading coefficient is irrational, then the sequence $(P(n))_{n \ge 0}$ is uniformly distributed modulo 1.
https://en.wikipedia.org/wiki/Equidistributed_sequence
In particular, the sequence $(n^m/(2\pi))_{n \ge 0}$ is uniformly distributed modulo 1, yielding the positive answer to your question.
