How to show $\exp(ix)=\cos x+i\sin x$ using $\exp(z)=\sum_{n=0}^{\infty}\frac{z^n}{n!}$ I have the following problem:

How to show $\exp(ix)=\cos x+i\sin x$ using $\exp(z)=\sum_{n=0}^{\infty}\frac{z^n}{n!}$

My attempt: 
\begin{align*}
\exp(ix)=\sum_{n=0}^{\infty}\frac{(ix)^n}{n!}
\end{align*}
Now split the sum in 4 sums because $i^n$ has period 4, more precisely $i^{4n}=1$, $i^{4n+1}=i$, $i^{4n+2}=-1$ and $i^{4n+3}=-i$ for $n\geq 0$.
\begin{align*}
\sum_{n=0}^{\infty}\frac{(ix)^n}{n!}&=\sum_{n=0}^{\infty}\frac{(ix)^{4n}}{(4n)!}+\sum_{n=0}^{\infty}\frac{(ix)^{4n+1}}{(4n+1)!}+\sum_{n=0}^{\infty}\frac{(ix)^{4n+2}}{(4n+2)!}+\sum_{n=0}^{\infty}\frac{(ix)^{4n+3}}{(4n+3)!}\\
&=\sum_{n=0}^{\infty}\left(\frac{x^{4n}}{(4n)!}-\frac{x^{4n+2}}{(4n+2)!}\right)+i\sum_{n=0}^{\infty}\left(\frac{x^{4n+1}}{(4n+1)!}-\frac{x^{4n+3}}{(4n+3)!}\right)
\end{align*}
Then $4n$ and $4n+2$ are nonnegative and even numbers and $4n+1$ and $4n+3$ are odd. How can I use this to convert the two last expressions into
\begin{align*}
\cos x&=\sum_{m=0}^{\infty}(-1)^m \frac{x^{2m}}{(2m)!}\\
\sin x&=\sum_{m=0}^{\infty}(-1)^m \frac{x^{2m+1}}{(2m+1)!}
\end{align*}
I don't want to see proofs like
\begin{align*}
\exp(ix)&=1+ix+\frac{(ix)^2}{2!}+\frac{(ix)^3}{3!}+\frac{(ix)^4}{4!}+\frac{(ix)^5}{5!}\ldots\\
&=\Big(1-\frac{x^2}{2!}+\frac{x^4}{4!}\mp\ldots\Big)+i\Big(x-\frac{x^3}{3!}+\frac{x^5}{5!}\mp\ldots\Big)\\
&=\cos x+i\sin x
\end{align*}


We obtain
\begin{align*}
\cos(x)&=\sum_{n=0}^{\infty}(-1)^n \frac{x^{2n}}{(2n)!}\\
&=\sum_{{n=0}\atop{n \text{ even}}}^{\infty}(-1)^n \frac{x^{2n}}{(2n)!}+\sum_{{n=0}\atop{n \text{ odd}}}^{\infty}(-1)^n \frac{x^{2n}}{(2n)!}\\
&=\sum_{{n=0}}^{\infty} \frac{x^{4n}}{(4n)!}-\sum_{{n=0}}^{\infty} \frac{x^{4n+2}}{(4n+2)!}\\
&=\sum_{{n=0}}^{\infty}\left( \frac{x^{4n}}{(4n)!}- \frac{x^{4n+2}}{(4n+2)!}\right)
\end{align*}
and similarly for $\sin(x)$.

