Prove $e^{z+w}=e^ze^w$ using the Taylor expansion for $ e^z.$ For all $z,w ∈ \Bbb{C}$, using the power series definition for $e^z$, prove $e^{z+w}=e^ze^w$.
 A: Proved it the following way:
Claim: $e^{z+w}=e^ze^w$, where $z\;and\;w\in \Bbb{C}$.
Proof: 
Let $e^z=1+z+\frac{z^2}{2!}+\frac{z^3}{3!}+\cdots$ 
and  $e^w=1+w+\frac{w^2}{2!}+\frac{w^3}{3!}+\cdots$ .
Now, for RHS:
$e^ze^w=\big(1+z+\frac{z^2}{2!}+\frac{z^3}{3!}+\cdots\big)\big(1+w+\frac{w^2}{2!}+\frac{w^3}{3!}+\cdots\big)$
$=1+w+\frac{w^2}{2!}+\frac{w^2}{3!}+z+zw+\frac{zw^2}{2!}+\frac{zw^3}{3!}+\cdots$
$=1+(z+w)+(\frac{z^2}{2!}+zw+\frac{w^2}{2!})+(\frac{z^3}{3!}+\frac{z^2w}{2!}+\frac{zw^2}{2!}+\frac{w^2}{3!})\cdots$
$=1+(z+w)+(\frac{z^2}{2!}+\frac{2zw}{2!}+\frac{w^2}{2!})+(\frac{z^3}{3!}+\frac{3z^2w}{3!}+\frac{3zw^2}{3!}+\frac{w^2}{3!})\cdots$
$=1+(z+w)+\frac{(z+w)^2}{2!}+\frac{(z+w)^3}{3!}+\cdots$
$=e^{z+w} \;QED.$
A: Starting with $e^{z+w}$, we have :
\begin{align}
e^{z+w} & = \sum_{n=0}^\infty \frac{(z+w)^n}{n!} \\ & = \sum_{n=0}^\infty \sum_{k=0}^n \binom{n}k \frac{z^kw^{n-k}}{n!}
\end{align}
Here's the point : we now need to collect terms of the form $z^K$ for fixed $K$. This will involve a change of variable.

To see how we do this, we write all the terms of this summation down , breaking it for $n = 0,1,2,...$:
\begin{align}
n = 0 & : & \binom{0}{0} \frac{z^0w^0}{ 0!} \\
n=1 & : & \binom{1}{0} \frac {z^0w^1}{ 1!} , \binom{1}{1} \frac{z^1w^0}{1!} \\
&\vdots& \\
n = N &:& \binom N0 \frac{z^0w^N}{N!} , \binom N1\frac{z^1 w^{N-1}}{N!}, \ldots , \binom NN \frac{z^Nw^0}{N!}\\
&\vdots&
\end{align}
This is one way of summing the terms : take all the terms for fixed $n$, sum these, and then sum over all $n$.
Now, let us do something different : we don't fix $n$. We fix $\mathbf k$. Now, let us write down what terms in the summation have $k=0$:
$$
\binom{0}{0} \frac{z^0w^0}{0!} , \binom 10 \frac{z^0w^1}{1!} , \binom 20 \frac {z^0w^2}{2!} , \ldots
$$
for say $k=7$, the $n = 0 \to 6$ will not contribute because $\binom nk$ in that case is zero. So if we look at which terms have $k=7$ in the summation, we get :
$$
\binom{7}{7} \frac{z^7w^0}{0!} + \binom 87 \frac{z^7w^1}{1!} + \binom 97 \frac{z^7w^2}{2!} + ...
$$
Now, the point is this : initially, we fixed $n$, and summed over all $k$, then summed over all $n$. Now, we fix $k$,sum over all $n$ then over all $k$.
Basically, I am trying to say that the summations can be interchanged :
$$
\sum_{n=0}^\infty \sum_{k=0}^n \binom{n}{k} \frac{z^kw^{n-k}}{k!} = \sum_{k=0}^\infty \color{green}{\sum_{n \geq k}} \binom {n}{k} \frac{z^kw^{n-k}}{n!}
$$
This is an inversion of question : first, we asked "Fix $n$, for which $k$ is the summation term non-zero?" for which the answer was $0 \leq k \leq n$. Now, we ask "fix $k$, for which $n$ is the summation term non-zero?" Now the answer should be clear : it is so only when $n\geq k$ (think about this yourself).
Now, we have fixed $k$. Note that this is equivalent to calculating the coefficient of $z^k$. We only need to sum over $n$. But as $k$ is fixed, so is $z^k$ and $k!$, these come out of the summation. So the rest of the answer follows.

So fixing $K$, look at the coefficient of $z^K$. for $n < K$ we have $\binom nK = 0$, so no coefficient. For $n \geq K$, the term becomes $\binom nK \frac{w^{n-K}}{n!}$ , which after simplifying the factorial becomes $\frac{w^{n-K}}{K!(n-K)!}$. 
In short, once we sum over all $n$, the coefficient of $z^K$ is :
$$
\sum_{n \geq K} \frac{w^{n-K}}{K!(n-K)!} \overbrace{=}^{t = n-K} \frac 1{K!}\sum_{t=0}^\infty\frac{w^t}{t!} = \frac{e^w}{K!}
$$
So now, once we have collected all the terms :
$$
e^{z+w} = \sum_{K=0}^\infty \frac {e^w}{K!} z^K = e^w \sum_{K=0}^\infty \frac{z^K}{K!} = e^we^z
$$
as expected.
A: Let $z,w \in \mathbb{C}$ be given. The defining series of $\exp(z)$ and $\exp(w)$ are both absolutely convergent, this means that their Cauchy-product converges to the product of their limit, i.e.:
\begin{align}
\exp(z)\exp(w)
&=\sum_{i=0}^{+\infty}\frac{z^i}{i!}\sum_{j=0}^{+\infty}\frac{w^j}{j!}\\
&=\sum_{i=0}^{+\infty}\sum_{k=0}^{i} \frac{z^k}{k!}\frac{w^{i-k}}{(i-k)!}\\
&=\sum_{i=0}^{+\infty}\sum_{k=0}^{i} z^k w^{i-k} \frac{1}{k! (i-k)!}\\
&=\sum_{i=0}^{+\infty}\frac{1}{i!}\sum_{k=0}^{i} z^k w^{i-k} \frac{i!}{k! (i-k)!}\\
&=\sum_{i=0}^{+\infty}\frac{(z+w)^i}{i!}\\
&=\exp(z+w)
\end{align}
A: From the definition of the exponential, you know that $f(x) = e^x$ verifies $f'(x) = f(x).$ Moreover it is clear that the solutions of this differential equation are all the $C e^x$ with $C$ a constant. Now, for fixed $w$, consider the function $f_w : z \mapsto e^{z+w}.$ One has, $$f_w'(z) = e^{z+w} = f_w(z).$$ Hence $$f_w(z) = C e^{z}$$ for some $C$. By evaluating at $z=0$, you will find $C=e^w$ and get the conclusion.
A: Knowing $e^{z}=1+z+z^2/2!+\cdots$ and $e^{w}=1+w+w^2/2!+\cdots,$
we have 
 $e^{z}.e^{w}=(1+z+z^2/2!+......) (1+w+w^2/2!+......)=1+(z+w)+(z^2+w^2+2zw)/2!+.......=1+(z+w)+(z+w)^2/2!+.......=e^{z+w}.$
Check the other few terms. 
