Prove that

$$e^{z+w} = \sum\_{k=0}^\infty \frac{z^k}{k!}\sum\_{m=0}^\infty \frac{w^m}{m!}=e^ze^w$$

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The proof provided is:

\begin{align}
\sum\_{n=0}^\infty \frac{(z+w)^n}{n!} &= \sum\_{n=0}^\infty \frac{1}{n!} \sum\_{k=0}^n \frac{n!}{k!(n-k)!} z^{n-k}w^k \\
&= \sum\_{n=0}^\infty \sum\_{k=0}^n \frac{z^{n-k}w^k}{k!(n-k)!} \tag{2} \\
&= \sum\_{k=0}^\infty \frac{w^k}{k!}\sum\_{n=0}^\infty \frac{z^{n-k}}{(n-k)!}\tag{3} \\
&= \sum\_{k=0}^\infty \frac{w^k}{k!}\sum\_{m=0}^\infty \frac{z^{m}}{m!} & (m=n-k)\tag{4} 
\end{align}

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Now from line $(2)$ to line $(3)$ why does the upper limit of the $k$ summation change to infinity? On the line $(4)$ why is the bottom limit of the inner sum not $m=-k$ as opposed to $m=0$?