TITLE: Show that $e^{x+y}=e^xe^y$ using $e^x=\lim_{n\to\infty }\left(1+\frac{x}{n}\right)^n$.
QUESTION [14 upvotes]: I was looking for a proof of $e^{x+y}=e^xe^y$ using the fact that $$e^x=\lim_{n\to\infty }\left(1+\frac{x}{n}\right)^n.$$
So I have that $$\left(1+\frac{x+y}{n}\right)^n=\sum_{k=0}^n\binom{n}{k}\frac{(x+y)^k}{n^k}=\sum_{k=0}^n\frac{1}{n^k}\sum_{i=0}^k\binom{k}{i}x^iy^{k-i}=\sum_{k=0}^n\binom{n}{k}\sum_{i=0}^k\binom{k}{i}\left(\frac{x}{n}\right)^i\left(\frac{y}{n}\right)^{k-i}$$
But I can't get $$\left(1+\frac{x+y}{n}\right)^n=\sum_{k=0}^n\binom{n}{k}\left(\frac{x}{n}\right)^k\sum_{i=0}^n\binom{n}{i}\left(\frac{y}{n}\right)^i.$$
Any idea ?

REPLY [6 votes]: \begin{align*}
e^x e^y &= \lim_{n\rightarrow \infty}(1+\frac{x}{n})^n\lim_{n\rightarrow \infty}(1+\frac{y}{n})^n\\
&=\lim_{n\rightarrow \infty}(1+\frac{x}{n})^n(1+\frac{y}{n})^n\\
&=\lim_{n\rightarrow \infty}(1+\frac{x+y}{n}+\frac{xy}{n^2})^n\\
&=\lim_{n\rightarrow \infty}(1+\frac{x+y}{n})^n \frac{(1+\frac{x+y}{n}+\frac{xy}{n^2})^n}{(1+\frac{x+y}{n})^n}\\
&=\lim_{n\rightarrow \infty}(1+\frac{x+y}{n})^n \left(1+\frac{xy}{n^2(1+\frac{x+y}{n})}\right)^n\\
&=\lim_{n\rightarrow \infty}(1+\frac{x+y}{n})^n\left(1+\frac{xy}{n^2(1+\frac{x+y}{n})}\right)^{\frac{n^2(1+\frac{x+y}{n})}{xy}\frac{xy}{n^2(1+\frac{x+y}{n})}n}\\
&= \lim_{n\rightarrow \infty}(1+\frac{x+y}{n})^n\\
&=e^{x+y}.
\end{align*}