Given that $\left\{\frac{\varphi}{2 \pi}\right\} \neq 0$, we have $e^{i \varphi} \neq 1$. By Lemma 3, we have
$$\begin{array}{l} \sum_{m=0}^{n-1} e^{i m \varphi}=\frac{1-e^{i n \varphi}}{1-e^{i \varphi}}=\frac{e^{\frac{i n \varphi}{2}}\left(e^{\frac{i n \varphi}{2}}-e^{-\frac{i n \varphi}{2}}\right)}{e^{\frac{i \varphi}{2}}\left(e^{\frac{i \varphi}{2}}-e^{-\frac{i \varphi}{2}}\right)}=e^{\frac{n \varphi \varphi}{2} \frac{i \varphi}{2}} \\ \quad \times \frac{\cos \frac{n \varphi}{2}+i \sin \frac{n \varphi}{2}-\cos \left(-\frac{n \varphi}{2}\right)-i \sin \left(-\frac{n \varphi}{2}\right)}{\cos \frac{\varphi}{2}+i \sin \frac{\varphi}{2}-\cos \left(-\frac{\varphi}{2}\right)-i \sin \left(-\frac{\varphi}{2}\right)} \\ \quad=e^{i \frac{n-1}{2} \varphi} \cdot \frac{\sin \frac{n \varphi}{2}}{\sin \frac{\varphi}{2}} \end{array}$$

Multiplying both sides of the above equation by $e^{i \theta}$, the lemma is proved.