$\left(\sum_{j=0}^\infty\frac{z^j}{j!}\right)\left(\sum_{k=0}^\infty\frac{w^k}{k!}\right)=\sum_{n=0}^\infty\sum_{j=0}^n\frac{z^jw^{n-j}}{j!(n-j)!}$ 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!
 A: This is an example of the change of order of summation (regrouping terms). Formally (without concern for convergence), the product
$$
\begin{align}
\left(\sum_{i=0}^\infty a_i\right)\left(\sum_{j=0}^\infty b_j\right)
&=\sum_{i=0}^\infty\sum_{j=0}^\infty a_ib_j\tag{1}\\
&=\sum_{k=0}^\infty\sum_{j=0}^k a_{k-j}b_j\tag{2}
\end{align}
$$
$(1)$ 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 $(2)$.
A: $$
(1+2+3+4+\cdots)\cdot\left(\begin{array} {} & \text{one} \\[6pt] + & \text{two} \\[6pt] + & \text{three} \\[6pt] + & \text{four} \\[6pt] + & \cdots \end{array}\right)
$$
$$
= \sum \left[ \begin{array}{cccc} 1\cdot\text{one}, & 2\cdot\text{one}, & 3\cdot\text{one}, & 4\cdot\text{one}, &  \cdots\\[6pt]
1\cdot\text{two}, & 2\cdot\text{two}, & 3\cdot\text{two}, & 4\cdot\text{two}, &  \cdots\\[6pt]
1\cdot\text{three}, & 2\cdot\text{three}, & 3\cdot\text{three}, & 4\cdot\text{three}, &  \cdots\\[6pt]
1\cdot\text{four}, & 2\cdot\text{four}, & 3\cdot\text{four}, & 4\cdot\text{four}, &  \cdots\\
\vdots & \vdots & \vdots & \vdots & \ddots
\end{array} \right]
$$
\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}
A: Multiplying the $j$ term on the left with the $k$ term on the right gives you
$\dfrac{z^j w^k}{j! k!}$.  Now if $j+k=n$, this is $\dfrac{z^j w^{n-j}}{j! (n-j)!}$.
Since $j$ and $k$ can be any nonnegative integers, the same is true for $n$.  Given $n$, $j$ can be any integer from $0$ to $n$.  
A: 
Definition: Given two series $ \sum_{n=0}^\infty a_n$ and
$\sum_{n=0}^\infty b_n$, define $$c_n=\sum_{k=0}^n a_{n-k}b_k.$$ The
series $\sum_{n=0}^\infty c_n$ is called Cauchy product of $\sum a_n$
and $\sum b_n$.

The Cauchy product is kinda discrete convolution. Check the following Theorem:

Mertens' theorem: If the series $\sum a_n$ and $\sum b_n$ converge to
$a$ and $b$, respectively, and, at least one of them converges
absolutely, then the Cauchy product of these two series converges to
$ab$.

In our case, by MacLaurin, $\sum_{n=0}^\infty \frac{t^n}{n!}$ converges absolutely, for $t\in\left\{z,w\right\}$ thus
$$\left(\sum_{n=0}^\infty \frac{z^n}{n!}\right)\left(\sum_{n=0}^\infty \frac{w^n}{n!}\right)=\sum_{n=0}^\infty c_n,$$
for $$c_n=\sum_{k=0}^n \frac{z^{n-k}}{(n-k)!}\frac{w^{k}}{k!}=\frac{1}{n!}\sum_{k=0}^n \frac{n!}{(n-k)!k!}z^{n-k}w^{k}=\frac{(z+w)^n}{n!}.$$
But then
$$e^ze^w=\left(\sum_{n=0}^\infty \frac{z^n}{n!}\right)\left(\sum_{n=0}^\infty \frac{w^n}{n!}\right)=\sum_{n=0}^\infty \frac{(z+w)^n}{n!}=e^{z+w},$$
as desired.
For more details check:

*

*Link: https://en.wikipedia.org/wiki/Cauchy_product#Convergence_and_Mertens'_theorem

*Book: Tom M. Apostol, Mathematical Analysis, Second Edition, page 204, Definition 8.45 and Theorem 8.46.

A: Given the series can be written
$=e^z e^w =e^{z+w}=\sum_{n=0}^\infty =\sum \sum....$
