How is it the complex exponential converges for any value of $z$ in the complex plane? $$e^{z} = 1 + \frac{z}{1!} + \frac{z^2}{2!} \cdots\cdots$$
How is it the "sum of exponents" rule holds for complex exponential, that is $e^{w}e^{z} = e^{w+z} $?
...using only the definition of $e^z$?
$$e^z=\sum_{n=0}^\infty\frac{z^n}{n!}$$ Putting $$a_n:=\frac{1}{n!}\Longrightarrow R:=\frac{1}{\lim_{n\to\infty}\sqrt[n]{|a_n|}}=\lim_{n\to\infty}\sqrt[n]{n!}=\infty$$
So the series has infinite convergence radius and is thus absolutely convergent for any $\,z\in\Bbb C\,$ , and from here it follows that $$e^{w+z}=\sum_{n=0}^\infty\frac{(w+z)^n}{n!}=\sum_{n=0}^\infty\sum_{k=0}^n\binom{n}{k}\frac{w^kz^{n-k}}{n!}=\sum_{n=0}^\infty\sum_{k=0}^n\frac{w^kz^{n-k}}{k!(n-k)!}=$$ $$=\sum_{k=0}^\infty\frac{w^k}{k!}\sum_{n=0}^\infty\frac{z^n}{n!}=:e^we^z$$