Here is the equation if you can't see it in the image: $\dfrac{e^{ni\theta}-1}{e^{i\theta}-1}=\dfrac{\sin\frac{n\theta}{2}}{\sin\frac{\theta}{2}}$
So in trying to prove this equation (using the left hand side) I'm reaching a certain point beyond which it seems impossible to go. It seems that the equation is incorrect.
Here is where I get to, and it seems easy enough to achieve. The left hand side comes out to be equal to $e^{\frac{(n-1)i\theta}{2}}\times \dfrac{\sin \frac{n\theta}{2}}{\sin \frac{\theta}{2}}$