trigonometry - How to prove theorem using Euler's formula?

I'm having a great deal of trouble with this proof.

"Prove $\cos θ + \cos 3θ + \cos 5θ + \cdots + \cos [(2n-1)θ] = \dfrac{\sin 2nθ}{2 \sin θ}$.
Prove $\sin θ + \sin 3θ + \sin 5θ + \cdots + \sin [(2n-1)θ] = \dfrac{(\sin nθ)^2}{\sin θ}$."

Relevant Equations

Euler's formula is $e^{iθ} = \cos θ + i \sin θ$.
The geometric progression formula is $S_n = a\left(\dfrac{1-r^n}{1-r}\right)$, where $a$ is the first term and $r$ is the constant that each term is multiplied by to get the next term.

My Attempts to Complete the Proof

Obviously, I need to use the geometric progression formula to prove this. With Euler's formula, the initial term $a$ is $e^{iθ}$. The $r$ term is $e^{2iθ}$. I believe these two values for $a$ and $r$ are correct, as the first term will simply be $\cos θ + i \sin θ$, the second will be $e^{iθ + 2iθ} = \cos 3θ + i \sin 3θ$, and so on.

When I plug this into the formula, I get $S_{2n-1} = e^{iθ}\left(\dfrac{1-e^{2iθ (2n-1)}}{1-e^{2iθ}}\right)$.

This is fine, but I can't for the life of me simplify this to anything meaningful. Did I make a mistake somewhere? Are these the correct values for $a$ and $r$? Is there some special trick that I'm missing?

ANY help would be appreciated. I've been stuck on this for days.

Thanks in advance,
Leo

EDIT: how do I get the last step? I end up with $\dfrac{1}{2} \dfrac{\sin (4n-2)θ}{2 \sin θ}$. It seems that the $2n-1$ is causing problems. If it was just $n$, then this would all work out perfectly. How can I fix this?