Prove that we have
$$\begin{aligned} \frac{1}{e^{2 \pi i \frac{1}{n}}-1} & =\frac{e^{-2 \pi i \frac{1}{n}}+1}{\left(e^{2 \pi i \frac{1}{n}}-1\right)\left(e^{-2 \pi i \frac{1}{n}}+1\right)} \\ & =\frac{1+\cos \left(-\frac{2 \pi}{n}\right)+i \sin \left(-\frac{2 \pi}{n}\right)}{e^{2 \pi i \frac{1}{n}}-e^{-2 \pi i \frac{1}{n}}} \\ & =\frac{1+\cos \frac{2 \pi}{n}-i \sin \frac{2 \pi}{n}}{2 i \sin \frac{2 \pi}{n}} \\ & =-\frac{1}{2}-\frac{1+\cos \frac{2 \pi}{n}}{2 \sin \frac{2 \pi}{n}} i \end{aligned}$$

By Lemma 7, we have
$$\begin{aligned} & \sum_{m=1}^{n-1} m e^{2 \pi i \frac{m}{n}}-\frac{n}{e^{2 \pi i \frac{1}{n}}-1}=0 \\ = & \sum_{m=1}^{n-1} m \cos \frac{2 \pi m}{n}+i \sum_{m=1}^{n-1} m \sin \frac{2 \pi m}{n} \\ & -n\left(-\frac{1}{2}-\left(\frac{1+\cos \frac{2 \pi}{n}}{2 \sin \frac{2 \pi}{n}}\right) i\right) \\ = & \sum_{m=1}^{n-1} m \cos \frac{2 \pi m}{n}+\frac{n}{2}+\left(\sum_{m=1}^{n-1} m \sin \frac{2 \pi m}{n}\right. \\ & \left.+\frac{\left(1+\cos \frac{2 \pi}{n}\right) n}{2 \sin \frac{2 \pi}{n}}\right) i=0, \text{ hence Lemma 8 is proved. } \end{aligned}$$