Prove properties of exponential function using a limit definition I am using ${\displaystyle \lim_{n \to \infty} \left( 1 + \frac{x}{n} \right)^{n}}=e^x$ and want to show that $e^{x+y} =e^xe^y$ and $e^{-x}=(e^x)^{-1}$ using only the given limit. 
The trouble is that I believe $e^{x+y}=e^xe^y$ is not true for large values of $n$
 A: Since
$$\left(1+\frac{x}{n}\right)^n\cdot\left(1+\frac{y}{n}\right)^n = \left(1+\frac{x+y}{n}+\frac{xy}{n^2}\right)^n=\left(1+\frac{x+y}{n}\right)^n\cdot\left(1+\frac{xy}{n^2+nx+ny}\right)^n$$
we just need to show that for any $x,y$ we have:
$$\lim_{n\to +\infty}\left(1+\frac{xy}{n^2+nx+ny}\right)^n=1\tag{1}$$
that follows from:
$$\lim_{n\to +\infty}\left(1+\frac{xy}{n^2}\right)^n=1.\tag{2}$$
A: Case 1:  Assume $x+y>0$
Write $e^{x+y}={\displaystyle \lim_{n \to \infty} \left( 1 + \frac{(x+y)}{n} \right)^{n}}$ and substitute $\frac1m=\frac{x+y}{n}$.
$$e^{x+y}={\displaystyle \lim_{n \to \infty} \left( 1 + \frac{(x+y)}{n} \right)^{n}}={\displaystyle \lim_{m \to \infty} \left( 1 + \frac{1}{m} \right)^{m(x+y)}}=\left({\displaystyle \lim_{m \to \infty} \left( 1 + \frac{1}{m} \right)^{m}}\right)^{x+y}=e^xe^y$$
which completes the proof for Case 1, $x+y>0$. 
Case 2: $x+y=0$.  This case for is trivial.
Case 3:  $x+y<0$
When $x+y<0$, we begin by making the substitution $\frac{1}{m} = -\frac{x+y}{n}$.  We proceed analogously thereafter by making use of the identity 
$$e^{-1}=\lim_{n \to \infty} \left(1-\frac{1}{n}\right)^n$$
