How complex exponential converges and "sum of exponents" rule holds 
*

*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 $e^z$?
 A: *

*One can use the ratio test or notice that for every $z \in \mathbb{C}$ one has $|z| \in \mathbb{R}$ and 
$$
\sum_{n=0}^\infty\frac{|z|^n}{n!}=e^{|z|},
$$
i.e. the series $\sum_{n=0}^\infty z^n/n!$ converges absolutely, and therefore it converges for every $z \in \mathbb{C}$.


2.
$$
e^we^z=(\sum_{n=0}^\infty\frac{w^n}{n!})(\sum_{n=0}^\infty\frac{z^n}{n!})=\sum_{n=0}^{\infty}A_n(w,z),
$$
where with the use of Cauchy product one has
\begin{eqnarray}
A_n(w,z)&=&\sum_{k=0}^n\frac{w^k}{k!}\frac{z^{n-k}}{(n-k)!}=\sum_{k=0}^n\frac{1}{n!}{n\choose k}w^kz^{n-k}=\frac{1}{n!}\sum_{k=0}^n{n\choose k}w^kz^{n-k}=\frac{(w+z)^n}{n!}.
\end{eqnarray}
A: $$e^z=\sum_{n=0}^\infty\frac{z^n}{n!}$$
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
$$e^{w+z}=\sum_{n=0}^\infty\frac{(w+z)^n}{n!}=\sum_{n=0}^\infty\sum_{k=0}^n\binom{n}{k}\frac{w^kz^{n-k}}{n!}=\sum_{n=0}^\infty\sum_{k=0}^n\frac{w^kz^{n-k}}{k!(n-k)!}=$$
$$=\sum_{k=0}^\infty\frac{w^k}{k!}\sum_{n=0}^\infty\frac{z^n}{n!}=:e^we^z$$
A: The second part can be proved also using the identity theorem: Two analytic functions are the same if they coincide on a set with a limit point.  Look at $e^{w+z}$ and $e^w e^z$  with $w$ a fixed real number. The two functions (as functions of $z$) coincide on the real line.  Therefore, they are equal for all $z \in \mathbb{C}$. Since $w$ was arbitrary they also coincide as functions of $w$ on the real line. Again by the identity theorem $e^{w+z}=e^w e^z$ for all $w \in \mathbb{C}$. 
