Limit of the sequence $(\sin n)^{n}$

How to calculate $$ \lim_{n\to\infty}(\sin n)^{n} \, ? $$ Is it sufficiently that since $|\sin x|\leq 1$ $\forall x\in\mathbb{R}$ and $|\sin n|<1$ $\forall n\in\mathbb{N}$ then $$ \lim_{n\to\infty}(\sin n)^{n}=0 \, ? $$ Is it true that if $|a_{n}|<1$ $\forall n\in\mathbb{N}$ then $$ \lim_{n\to\infty}(a_{n})^{n}=0 \, ? $$

This limit does not exist, you can find two distinct accumulation points.

By the theorem of Dirichlet, you will find as many integers $p,q$ as you want such that $$|\pi-\frac pq|<\frac1{q^2},$$or $$|q\pi-p|<\frac1{q}.$$ Taking the sine and raising to the $p^{th}$ power, $$|\sin^pp|<\sin^p\frac1{q},$$ i.e. with $q>\frac p4$, $\color{blue}{\sin^pp}$ is arbitrarily close to $\color{blue}0$ infinitely many times.

For the same reason, you will find as many integers $p,q=2^er$ (odd $r$) as you want such that $$|\frac\pi{2^{e+1}}-\frac pq|<\frac1{q^2},$$or $$|r\frac\pi{2}-p|<\frac1{q}.$$ Taking the cosine (a decreasing function) and raising to the $p^{th}$ power, $$|\sin^pp|>\cos^p\frac1{q},$$ i.e. noting that $q>2^{e-1}p$, $\color{blue}{\sin^pp}$ is arbitrarily close to $\color{blue}1$ infinitely many times (as $\cos\frac1q=1-o\left(\frac1q\right)$).

Numerical calculations strongly suggest that the sequence $(\sin n)^n$ is divergent. Here is the plot of the first million terms: