Prove $e^{x+y}=e^{x}e^{y}$ by using Exponential Series

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: $$ \begin{array}{ccccccccc} & & & & & & & & n=3 \\ & & & & & & & \swarrow \\ a_{0,0} & & a_{0,1} & & a_{0,2} & & a_{0,3} & & \cdots \\ \\ & & & & & \swarrow \\ a_{1,0} & & a_{1,1} & & a_{1,2} & & a_{1,3} & & \cdots \\ \\ & & & \swarrow \\ a_{2,0} & & a_{2,1} & & a_{2,2} & & a_{2,3} & & \cdots \\ & \swarrow \\ a_{3,0} & & a_{3,1} & & a_{3,2} & & a_{3,3} & & \cdots \\ \vdots & & \vdots & & \vdots & & \vdots \end{array} $$

Try to expand $\displaystyle\frac{(x+y)^n}{n!}$.

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.