Limit of $\lim_{n\to\infty}n\cdot \sqrt[n]{\sin\left(\frac{1}{1}\right)\cdot \sin\left(\frac{1}{2}\right)\cdots\sin\left(\frac{1}{n}\right)}$ $$\lim _{n\to \infty }\left(n\cdot \sqrt[n]{\sin\left(\frac{1}{1}\right)\cdot \sin\left(\frac{1}{2}\right)\cdot \cdot \cdot \sin\left(\frac{1}{n}\right)}\right)$$
I have noticed that the nth root is in fact the nth term in the series of geometric means of the series $ \sin\left(\frac{1}{n}\right)$.
Therefore i believe that the series behaves like $ \frac{\sin\left(\frac{1}{n}\right)}{\frac{1}{n}}$ and therefore approaches 1. but i cant find a way of proving this.
I see that for each n we have $ \frac{\sin\left(\frac{1}{n}\right)}{\frac{1}{n}} \le \left(n\cdot \sqrt[n]{\sin\left(\frac{1}{1}\right)\cdot \sin\left(\frac{1}{2}\right)\cdot \cdot \cdot \sin\left(\frac{1}{n}\right)}\right)$ and that this implies that the limit is greater than or equal to 1 but cant get much further than this.
This problem is meant to be solved without the notion of continuity and without certainly without taylor series expansion, lhospitals rule etc..
Any help would be appreciated. Thanks!

$$\begin{align}\lim _{n\to \infty }\left(n\cdot \sqrt[n]{\sin\left(\frac{1}{1}\right)\cdot \sin\left(\frac{1}{2}\right)\cdot \cdot \cdot \sin\left(\frac{1}{n}\right)}\right)&=\lim _{n\to \infty }\left(n\cdot \sqrt[n]{\prod_{i=1}^n \sin\dfrac1i}\right) &\\=& \lim _{n\to \infty }\left(n\cdot \sqrt[n]{\prod_{i=1}^n \sin\dfrac1i}\right) &\\=& \lim _{n\to \infty }\left(\dfrac{n}{\sqrt[n]{n!}}\cdot \sqrt[n]{  \prod_{i=1}^n \dfrac{\sin (1/i)}{1/i}}\right)&\\=& \lim _{n\to \infty }\dfrac{n}{\sqrt[n]{n!}}\cdot\lim_{n\to \infty} \sqrt[n]{  \prod_{i=1}^n \dfrac{\sin (1/i)}{1/i}}&\\=& e\cdot\exp\left(\lim_{n\to\infty}\dfrac1n\sum^n_{i=1}\ln\left(\dfrac{\sin (1/i)}{1/i}\right)\right)&\\=& e\cdot\exp\left(0\right)=e \end{align}$$
I used Finding the limit of $\frac {n}{\sqrt[n]{n!}}$ and Convergence and value of infinite product $\prod^{\infty}_{n=1} n \sin \left( \frac{1}{n} \right)$?.
