Proving multiplicative property of the exponential function from its limit definition I am trying to show the function defined by:
$$E(x)=\lim_{n\to\infty}\left(1+\frac{x}{n}\right)^n$$
satisfies the property:
$$E(x)E(y)=E(x+y)$$

Assuming $E$ is well defined, I can interchange products and limits (?). We have:
$$\begin{aligned} E(x)E(y) &=\lim_{n\to\infty}\left(1+\frac{x}{n}\right)^n\lim_{n\to\infty}\left(1+\frac{y}{n}\right)^n \\ &=\lim_{n\to\infty}\left[\left(1+\frac{x}{n}\right)\left(1+\frac{y}{n}\right)\right]^n \\ &=\lim_{n\to\infty}\left[1+\frac{x+y}{n}+\frac{xy}{n^2}\right]^n\\ &=\lim_{n\to\infty} \sum_{k=0}^n\binom{n}{k}\left(1+\frac{x+y}{n}\right)^{n-k}\left(\frac{xy}{n^2}\right)^k
\\ & = E(x+y)+\lim_{n\to\infty} \sum_{k=1}^n\binom{n}{k}\left(1+\frac{x+y}{n}\right)^{n-k}\left(\frac{xy}{n^2}\right)^k\end{aligned}$$
This is where I am stuck. I can see that, for any $k$:
$$\binom{n}{k}\left(1+\frac{x+y}{n}\right)^{n-k}\left(\frac{xy}{n^2}\right)^k=\left(\frac{xy}{n}\right)^k\left(1+\frac{x+y}{n}\right)^{n-k}\prod_{r=0}^{k-1}\left(1-\frac{r}{n}\right)\overset{n\to\infty}\to 0$$
But because the sum increases in terms as $n$ increases, I feel like this is not sufficient to argue that the limit of it is $0$. Is that right? If so I'm not sure how to go about it and would appreciate some help.
 A: You do not need to deal with that hard expansion. Here is another way using Squeeze theorem for $xy\ge 0$. The case $xy\le 0$ is also similar. Notice that for any $a>0$ and large enough $n$ we have:$$1+{x+y\over n}\le 1+{x+y\over n}+{xy\over n^2}<1+{x+y\over n}+{xy\over an}=1+{x+y+{xy\over a}\over n}$$using Squeeze theorem we have$$\lim _{n\to \infty}(1+{x+y\over n})^n\le \lim _{n\to \infty}(1+{x+y\over n}+{xy\over n^2})^n\le \lim _{n\to \infty}(1+{x+y+{xy\over a}\over n})^n$$or using the definition $$E(x+y)\le E(x)E(y)\le E(x+y+{xy\over a})$$since this is true for any $a>0$ by tending $a$ to $\infty$ we obtain $$E(x+y)\le E(x)E(y)\le E(x+y)$$which yields to $$E(x+y)=E(x)E(y)$$
A: I don't recommend trying to do it this way. The cleanest proof I know is a bit less direct than this. First, show that $E(x)$ is the unique solution to the differential equation $E'(x) = E(x)$ with initial condition $E(0) = 1$, and more generally that $C E(x)$ is the unique solution with initial condition $E(0) = C$. 
For existence you'll want to exchange a limit and a derivative and you'll need to be careful about that, but once you've justified that exchange, $\frac{d}{dx} \left( 1 + \frac{x}{n} \right)^n = \left( 1 + \frac{x}{n} \right)^{n-1}$ so it's clear that the two limits are the same. Morally the point is that the limit definition of $E(x)$ is attempting to solve this differential equation using Euler's method. 
Uniqueness is easier: if $E_1(x)$ and $E_2(x)$ are two solutions compute the derivative of $\frac{E_1(x)}{E_2(x)}$. (Well, first show that solutions are always positive, so we can take this quotient.)
Once you have this everything is very easy: $E(x) E(y)$ and $E(x + y)$, as functions of $x$ with $y$ fixed, are both solutions to the differential equation $f'(x) = f(x)$ with initial condition $f(0) = E(y)$. And then we're done by uniqueness. 
A: In the excellent book
Analysis 1. 6., korrigierte Aufl. (German) 
Springer-Lehrbuch. Berlin: Springer. xiv, 398 S. (2001)
on page 78 (Exercise 14.) you find the following (approximately translated):
The exponential function as the limit of $\left(1+\frac x n\right)^n$.
Show that $E(x)=\lim E_n(x)$ exist, where $E_n(x)=\left(1+\frac x n\right)^n$, exists and that this limit equals $e^x$.
Hint: Following example 4.7 [Existence of $E_n(x)$, proved by application of the AGM inequality; AGM inequality=inequality between arithmetic and geometric mean] the sequence is (finally) monotonically increasing, and  $\left(1+\frac p n\right)\leq\left(1+\frac1 n\right)^p$ implies $E_n(p)\leq e^p$, i.e., $E_n(x)\leq E_n(y)\leq e^p$ for $-n\leq x\leq y\leq p$. Using the AGM inequality implies
$$ \left(1+\frac x n\right)^n \left(1+\frac y n\right)^n\leq \left(1+\frac {x+y} {2n}\right)^{2n}$$ and
$$\left(1+\frac x {n-1}\right)^{n-1} \left(1+\frac {xy} n\right)\leq \left(1+\frac x n+\frac y n+\frac{xy}{n^2}\right)^n=\left(1+\frac x n\right)^n\left(1+\frac y n\right)^n$$
implying $E(x)E(y)\leq E(x+y)\leq E(x)E(y)$
