Prove $e^{z+w}=e^ze^w$ using the Taylor expansion for $ e^z.$ For all $z,w ∈ \Bbb{C}$, using the power series definition for $e^z$, prove $e^{z+w}=e^ze^w$.

Let $z,w \in \mathbb{C}$ be given. The defining series of $\exp(z)$ and $\exp(w)$ are both absolutely convergent, this means that their Cauchy-product converges to the product of their limit, i.e.:
\begin{align}
\exp(z)\exp(w)
&=\sum_{i=0}^{+\infty}\frac{z^i}{i!}\sum_{j=0}^{+\infty}\frac{w^j}{j!}\\
&=\sum_{i=0}^{+\infty}\sum_{k=0}^{i} \frac{z^k}{k!}\frac{w^{i-k}}{(i-k)!}\\
&=\sum_{i=0}^{+\infty}\sum_{k=0}^{i} z^k w^{i-k} \frac{1}{k! (i-k)!}\\
&=\sum_{i=0}^{+\infty}\frac{1}{i!}\sum_{k=0}^{i} z^k w^{i-k} \frac{i!}{k! (i-k)!}\\
&=\sum_{i=0}^{+\infty}\frac{(z+w)^i}{i!}\\
&=\exp(z+w)
\end{align}
