Prove $e^{x+y}=e^{x}e^{y}$ by using Exponential Series In order to show $e^{x+y}=e^{x}e^{y}$ by using Exponential Series, I got the following: 
$$e^{x}e^{y}=\Big(\sum_{n=0}^{\infty}{x^n \over n!}\Big)\cdot \Big(\sum_{n=0}^{\infty}{y^n \over n!}\Big)=\sum_{n=0}^{\infty}\sum_{k=0}^n{x^ky^n \over {k!n!}}$$
But, where should I go next to get $e^{x+y}=\sum_{n=0}^{\infty}{(x+y)^n \over n!}$.
Thanks in advance. 
 A: Try to expand $\displaystyle\frac{(x+y)^n}{n!}$.
A: Let $f(t)=e^t$, as defined by the power series. Then by differentiating term by term, we find that $f'(t)=e^t$. Now let $y$ be fixed, and let
$$g_y(x)=\frac{f(x+y)}{f(x)}.$$
Differentiate. We get $g_y'(x)=0$, so $g_y$ is a constant function. Now let $x=0$. We find that $g_y(x)=e^y$, and the result follows.
A: You should have gotten
$$
\sum_{n=0}^\infty \sum_{k=0}^n \frac{x^k}{k!}\cdot\frac{y^{n-k}}{(n-k)!}.
$$
After that, you can write
$$
\sum_{n=0}^\infty \sum_{k=0}^n \frac{1}{n!} \cdot \frac{n!}{k!(n-k)!} x^k y^{n-k}.
$$
Since the factor $\dfrac{1}{n!}$ does not depend on $k$, you can pull it out:
$$
\sum_{n=0}^\infty \left(\frac{1}{n!} \sum_{k=0}^n \frac{n!}{k!(n-k)!} x^k y^{n-k}\right).
$$
Then you have
$$
\sum_{n=0}^\infty \frac{1}{n!} (x+y)^n.
$$
How does $\displaystyle\left(\sum_{n=0}^\infty b_n\right) \left(\sum_{m=0}^\infty c_m\right)$ become $\displaystyle\sum_{n=0}^\infty \sum_{m=0}^\infty (b_n c_m)$?
And how does $\displaystyle\sum_{n=0}^\infty \sum_{m=0}^\infty a_{n,m}$ become $\displaystyle\sum_{n=0}^\infty \sum_{k=0}^n a_{k,n-k}$?
In the first sum above, notice that $\sum_{m=0}^\infty c_m$ does not depend on $n$ so it can be pushed inside the other sum and become
$$
\sum_{n=0}^\infty \left( b_n \sum_{m=0}^\infty c_m \right).
$$
Then the factor $b_n$ does not depend on $m$, so that expression becomes
$$
\sum_{n=0}^\infty \sum_{m=0}^\infty (b_n c_m).
$$
That answers the first bolded question above.
Next consider the array
$$
\begin{array}{ccccccccc}
a_{0,0} & a_{0,1} & a_{0,2} & a_{0,3} & \cdots \\
a_{1,0} & a_{1,1} & a_{1,2} & a_{1,3} & \cdots \\
a_{2,0} & a_{2,1} & a_{2,2} & a_{2,3} & \cdots \\
a_{3,0} & a_{3,1} & a_{3,2} & a_{3,3} & \cdots \\
\vdots & \vdots & \vdots & \vdots
\end{array}
$$
The sum $\displaystyle\sum_{n=0}^\infty \sum_{k=0}^n a_{k,n-k}$ runs down diagonals:
$$
\begin{array}{ccccccccc}
& & & & & & & & n=3 \\
& & & & & & & \swarrow \\
a_{0,0} &  & a_{0,1} & & a_{0,2} & & a_{0,3} & & \cdots \\  \\
& & & & & \swarrow \\
a_{1,0} & & a_{1,1} & & a_{1,2} & & a_{1,3} & & \cdots \\  \\
& & & \swarrow \\
a_{2,0} & & a_{2,1} & & a_{2,2} & & a_{2,3} & & \cdots \\
& \swarrow \\
a_{3,0} & & a_{3,1} & & a_{3,2} & & a_{3,3} & & \cdots \\
\vdots & & \vdots & & \vdots & & \vdots
\end{array}
$$
