A popular way to prove this is to use the Taylor series for sin(x) and cos(x).
Adding them then gives the series for \(\displaystyle e^{ix}\)
\(\displaystyle \displaystyle \sum_{k=0}^{\infty}\frac{(-1)^{k}x^{2k}}{(2k)!}=cos(x)\)
\(\displaystyle \displaystyle \sum_{k=0}^{\infty}\frac{(-1)^{k}x^{2k+1}}{(2k+1)!}=sin(x)\)
Of course, multiply the sin series by i. Add them together and you should get:
\(\displaystyle e^{ix}=\displaystyle \sum_{k=0}^{\infty}\frac{(ix)^{k}}{k!}\)
Remember, \(\displaystyle i^{k}\) alternates.
\(\displaystyle i^{2}=-1, \;\ i^{3}=-i, \;\ i^{4}=1, \;\ i^{5}=i, \;\ i^{6}=-1\) and so on.