Csdrhrt

I am trying to show the function defined by:
$$E(x)=lim_{ntoinfty}left(1+frac{x}{n}right)^n$$
satisfies the property:
$$E(x)E(y)=E(x+y)$$

Assuming $E$ is well defined, I can interchange products and limits (?). We have:

$$begin{aligned} E(x)E(y) &=lim_{ntoinfty}left(1+frac{x}{n}right)^nlim_{ntoinfty}left(1+frac{y}{n}right)^n \ &=lim_{ntoinfty}left[left(1+frac{x}{n}right)left(1+frac{y}{n}right)right]^n \ &=lim_{ntoinfty}left[1+frac{x+y}{n}+frac{xy}{n^2}right]^n\ &=lim_{ntoinfty} sum_{k=0}^nbinom{n}{k}left(1+frac{x+y}{n}right)^{n-k}left(frac{xy}{n^2}right)^k
\ & = E(x+y)+lim_{ntoinfty} sum_{k=1}^nbinom{n}{k}left(1+frac{x+y}{n}right)^{n-k}left(frac{xy}{n^2}right)^kend{aligned}$$
This is where I am stuck. I can see that, for any $k$:
$$binom{n}{k}left(1+frac{x+y}{n}right)^{n-k}left(frac{xy}{n^2}right)^k=left(frac{xy}{n}right)^kleft(1+frac{x+y}{n}right)^{n-k}prod_{r=0}^{k-1}left(1-frac{r}{n}right)overset{ntoinfty}to 0$$
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.

You do not need to deal with that hard expansion. Here is another way using Squeeze theorem for $xyge 0$. The case $xyle 0$ is also similar. Notice that for any $a>0$ and large enough $n$ we have:$$1+{x+yover n}le 1+{x+yover n}+{xyover n^2}<1+{x+yover n}+{xyover an}=1+{x+y+{xyover a}over n}$$using Squeeze theorem we have$$lim _{nto infty}(1+{x+yover n})^nle lim _{nto infty}(1+{x+yover n}+{xyover n^2})^nle lim _{nto infty}(1+{x+y+{xyover a}over n})^n$$or using the definition $$E(x+y)le E(x)E(y)le E(x+y+{xyover a})$$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 $$E(x+y)=E(x)E(y)$$

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.

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 $left(1+frac x nright)^n$.
Show that $E(x)=lim E_n(x)$ exist, where $E_n(x)=left(1+frac x nright)^n$, 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 $left(1+frac p nright)leqleft(1+frac1 nright)^p$ implies $E_n(p)leq e^p$, i.e., $E_n(x)leq E_n(y)leq e^p$ for $-nleq xleq yleq p$. Using the AGM inequality implies
$$ left(1+frac x nright)^n left(1+frac y nright)^nleq left(1+frac {x+y} {2n}right)^{2n}$$ and

$$left(1+frac x {n-1}right)^{n-1} left(1+frac {xy} nright)leq left(1+frac x n+frac y n+frac{xy}{n^2}right)^n=left(1+frac x nright)^nleft(1+frac y nright)^n$$
implying $E(x)E(y)leq E(x+y)leq E(x)E(y)$