Could you help me how to find the limit of $$\left(\sin \frac{1}{n} \cdot \sin \frac{2}{n} \cdots \sin 1\right)^{\frac{1}{n}}?$$
I know that $$\ln \left((\sin \frac{1}{n} \cdot \sin \frac{2}{n} \cdots \sin 1)^{\frac{1}{n}} \right)=\frac{1}{n} \sum_{k=1}^n \ln \left( \sin(\frac{k}{n})\right)$$
and $$\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{k=1}^n \ln \left( \sin(\frac{k}{n})\right) = \int_0^1 \ln(\sin(x)) \, dx \text{ (Riemann integral)}$$
but I am not sure what to do next, I mean, how do I get back to $$\lim_{n \rightarrow \infty} (\sin \frac{1}{n} \cdot \sin \frac{2}{n} \cdots \sin 1)^{\frac{1}{n}}?$$
Could you help me with that?