How to obtain this limit Can you calculate rigorously the limit $\lim\limits_{n \to \infty} {(\sin n)^{\frac{1}{n}}}$

Following PEV's hint we argue as follows: There is a $p$ (e.g. $p=42$, according to this: Link) and a $k_0$ with $|\pi -{n\over k}|\geq 1/ k^p$ for all $k>k_0$ and all $n$. Assume $n>4k_0$ and let $k$ be the nearest integer to ${n\over \pi}$. Then $k>k_0$ and therefore $|n - k\pi|\geq 1/ k^{p-1}\geq C/ n^{p-1}$ for some $C>0$ which does not depend on $n$. As $|\sin(x)|\geq 2|x|/\pi$ $\ (|x|\leq{\pi\over2})$ it follows that $|\sin(n)|\geq C'/ n^{p-1}$. Since this is true for all large $n$ the indicated limit (with $|\sin|$ instead of $\sin$) is indeed $1$.
