<s>[INST] $\left(\sum_{j=0}^\infty\frac{z^j}{j!}\right)\left(\sum_{k=0}^\infty\frac{w^k}{k!}\right)=\sum_{n=0}^\infty\sum_{j=0}^n\frac{z^jw^{n-j}}{j!(n-j)!}$ 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!
 [/INST] $$
(1+2+3+4+\cdots)\cdot\left(\begin{array} {} & \text{one} \\[6pt] + & \text{two} \\[6pt] + & \text{three} \\[6pt] + & \text{four} \\[6pt] + & \cdots \end{array}\right)
$$
$$
= \sum \left[ \begin{array}{cccc} 1\cdot\text{one}, & 2\cdot\text{one}, & 3\cdot\text{one}, & 4\cdot\text{one}, &  \cdots\\[6pt]
1\cdot\text{two}, & 2\cdot\text{two}, & 3\cdot\text{two}, & 4\cdot\text{two}, &  \cdots\\[6pt]
1\cdot\text{three}, & 2\cdot\text{three}, & 3\cdot\text{three}, & 4\cdot\text{three}, &  \cdots\\[6pt]
1\cdot\text{four}, & 2\cdot\text{four}, & 3\cdot\text{four}, & 4\cdot\text{four}, &  \cdots\\
\vdots & \vdots & \vdots & \vdots & \ddots
\end{array} \right]
$$
\begin{align}
& = \cdots\cdots\cdots +\sum\left[ \begin{array}{cccc} \cdot & \cdot & 3\cdot\text{one}, & \cdot &  \cdots\\[6pt]
\cdot & 2\cdot\text{two}, & \cdot & \cdot &  \cdots\\[6pt]
1\cdot\text{three}, & \cdot & \cdot & \cdot &  \cdots\\[6pt]
\cdot & \cdot & \cdot & \cdot &  \cdots\\
\vdots & \vdots & \vdots & \vdots & \ddots
\end{array} \right] \\[18pt]
& {}\qquad\qquad\qquad{}+ \sum\left[ \begin{array}{cccc} \cdot & \cdot & \cdot & 4\cdot\text{one}, &  \cdots\\[6pt]
\cdot & \cdot & 3\cdot\text{two}, & \cdot &  \cdots\\[6pt]
\cdot & 2\cdot\text{three}, & \cdot & \cdot &  \cdots\\[6pt]
1\cdot\text{four}, & \cdot & \cdot & \cdot &  \cdots\\
\vdots & \vdots & \vdots & \vdots & \ddots
\end{array} \right] + \cdots\cdots\cdots
\end{align}
</s>