$\left(\sum_{j=0}^\infty\frac{z^j}{j!}\right)\left(\sum_{k=0}^\infty\frac{w^k}{k!}\right)=\sum_{n=0}^\infty\sum_{j=0}^n\frac{z^jw^{n-j}}{j!(n-j)!}$

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 $(2)$.

$$ (1+2+3+4+\cdots)\cdot\left(\begin{array} {} & \text{one} \\[6pt] + & \text{two} \\[6pt] + & \text{three} \\[6pt] + & \text{four} \\[6pt] + & \cdots \end{array}\right) $$ $$ = \sum \left[ \begin{array}{cccc} 1\cdot\text{one}, & 2\cdot\text{one}, & 3\cdot\text{one}, & 4\cdot\text{one}, & \cdots\\[6pt] 1\cdot\text{two}, & 2\cdot\text{two}, & 3\cdot\text{two}, & 4\cdot\text{two}, & \cdots\\[6pt] 1\cdot\text{three}, & 2\cdot\text{three}, & 3\cdot\text{three}, & 4\cdot\text{three}, & \cdots\\[6pt] 1\cdot\text{four}, & 2\cdot\text{four}, & 3\cdot\text{four}, & 4\cdot\text{four}, & \cdots\\ \vdots & \vdots & \vdots & \vdots & \ddots \end{array} \right] $$ \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}