TITLE: Prove $e^{x+y}=e^{x}e^{y}$ by using Exponential Series
QUESTION [16 upvotes]: 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.

REPLY [3 votes]: 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.