How do we know Taylor's Series works with complex numbers?

The series $f(z)=\sum_{n=0}^{\infty} \frac{z^n}{n!}$ is convergent on $C$, and thus it defines an Analytic function.

Now there are few different ways to convince yourself that this has to be $e^z$.

For once, it is the only Analytic continuation of $e^x$ to the complex plane...

Or, alternately, you can prove that $f(z_1+z_2)=f(z_1)f(z_2)$ and $f(1)=e$. Also, you can show that it is the only differentiable function satisfying these two relations.

If you prefer differential equations, $f'(z)=f(z)$ and $f(1)=e$ uniquely determine a solution, and bot $e^z$ and $f(z)$ are solutions....

You define a complex function by the formula $f(z)=\sum_{n=0}^{\infty} \frac{z^n}{n!} $. You prove that it converges everywhere and defines a holomorphic function of $z$. Then you prove that for $z=x \in \mathbb{R}$ it agrees with the usuall exponential.