$\left(\sum_{j=0}^\infty\frac{z^j}{j!}\right)\left(\sum_{k=0}^\infty\frac{w^k}{k!}\right)=\sum_{n=0}^\infty\sum_{j=0}^n\frac{z^jw^{n-j}}{j!(n-j)!}$ I've been going through some series notes from my lecture and got stuck at this equality: $$\left(\sum_{j=0}^\infty\frac{z^j}{j!}\right)\left(\sum_{k=0}^\infty\frac{w^k}{k!}\right)=\sum_{n=0}^\infty\sum_{j=0}^n\frac{z^jw^{n-j}}{j!(n-j)!}$$
Where $z,w\in\mathbb C$.

Rather than a proof, I'm looking more for a way of understanding this equality, so next time I see something similar I go "oh, right, I know that". Right now I only have vague ideas of why that would be true. 

Thanks a bunch for any help!
This is an example of the change of order of summation (regrouping terms). Formally (without concern for convergence), the product
$$
\begin{align}
\left(\sum_{i=0}^\infty a_i\right)\left(\sum_{j=0}^\infty b_j\right)
&amp;=\sum_{i=0}^\infty\sum_{j=0}^\infty a_ib_j\tag{1}\\
&amp;=\sum_{k=0}^\infty\sum_{j=0}^k a_{k-j}b_j\tag{2}
\end{align}
$$
$(1)$ is the distributing multiplication over addition.

$(2)$ is a change of variables: $i+j=k$

Each product in $(1)$ appears once and once only in $(2)$.
Given the series can be written

$=e^z e^w =e^{z+w}=\sum_{n=0}^\infty =\sum \sum....$