complex analysis - Prove the taylor series of $cos(2z)$

First i turned $$\cos(2z) = \frac{e^{2iz} + e^{-2iz}}{2}$$ , then using the taylor series of $$e^{z}$$ I calculated the taylor series of both arguments.
$$\frac{e^{2iz}}{2} = \sum_{n=0}^{\infty }\frac{(i)^{n}2^{n-1}z^{n}}{n!}$$
$$\frac{e^{-2iz}}{2}= \sum_{n=0}^{\infty }\frac{(-i)^{n}2^{n-1}z^{n}}{n!}$$
From this stage i dont know how to continue.
Is my way to the solution is right?

Answer

Hint:
$$(i)^n+(-i)^n=e^{\frac{in\pi}{2}}+e^{\frac{-in\pi}{2}}=2\cos{\dfrac{n\pi}{2}}.$$