Prove $e^{z+w}=e^ze^w$ for some $z,w\in\mathbb{C}$ by power series. Prove that
$$e^{z+w} = \sum_{k=0}^\infty \frac{z^k}{k!}\sum_{m=0}^\infty \frac{w^m}{m!}=e^ze^w$$

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}

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 $m=0$?

Note that for $k \ge n$ it is $\frac{1}{(n - k)!} = 0$.
Or go the other way around:
$\begin{align}
\mathrm{e}^w \cdot \mathrm{e}^z
   &= \sum_{r \ge 0} \frac{w^r}{r!} \cdot \sum_{s \ge 0} \frac{z^s}{s!} \\
   &= \sum_{n \ge 0} \sum_{0 \le k \le n} \frac{w^k z^{n - k}}{k! (n - k)!} \\
   &= \sum_{n \ge 0} \frac{1}{n!} 
         \sum_{0 \le k \le n} \frac{n!}{k! (n - k)!} w^k z^{n - k} \\
   &= \sum_{n \ge 0} \frac{1}{n!}
         \sum_{0 \le k \le n} \binom{n}{k} w^k z^{n - k} \\
   &= \sum_{n \ge 0} \frac{1}{n!} (w + z)^n \\
   &= \mathrm{e}^{w + z}
\end{align}$
Here we use first the Cauchy product of series (i.e., gather all terms with exponent of $w$ and $z$ adding up to $n$), and later the binomial theorem.
