I want to prove that $e^{x+y}=e^x\*e^y$, where x and y are complex numbers. I only want to use that evey complex number is on the form $z=a+bi$, where $i^2=-1$.
And if z is a complex number we have $e^z=\Sigma\_{n=0}^\infty\frac{z^n}{n!}$.

However I seem to get stuck, here is my attempt:

$x=a+bi$

$y=c+di$

$e^{x+y}=\Sigma\_{n=0}^\infty\frac{(a+bi\*c+di)^n}{n!}$. **I am not sure if we can use the binomial theorem on the part inside the summation, because we have $i$ there?, can we?**
If we can, I get

$=\Sigma\_{n=0}^\infty[\Sigma\_{k=0}^n\frac{n!}{k!(n-k)!}(a+bi)^k\*(c+di)^{n-k}\*\frac{1}{n!}]$. My problem is that I am not able to sepearte the n and the k, if I was able to show that this was equal to:

$\Sigma\_{n=0}^\infty\frac{(a+bi)^n}{n!}\*\Sigma\_{k=0}^\infty\frac{(c+di)^k}{k!}$ then I would have provedwhat I wanted, is there a way to get here?