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 $n$
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:
$$\lim_{n\to +\infty}\left(1+\frac{xy}{n^2}\right)^n=1.\tag{2}$$
At $x$ units, the value of the function is $e^x$. What happens when we move $y$ units furthur? The value then is $e^{x+y}$.

Divide the distance between $x$ and $x+y$ into $n$ intervals where $n$ is very large. The length of each interval is $\frac{y}{n}$. When we move from $x$ to $x+\frac{y}{n}$, what is the new value of the function?

You can think of $e^x$ as the principal money deposited in the bank. To get the new value, just multiply the principal money by $1+\frac{y}{n}$. This works because the exponential function is its own derivative. This is simply using the equation $e^{x+dx}=e^x+\frac{d(e^x)}{dx}dx=e^x(1+dx)$.

So the new value is $e^x(1+\frac{y}{n})$. This new value will work as our new principal money. When we furthur move by $\frac{y}{n}$ units, just multiply the new principal money by $1+\frac{y}{n}$ to get the next principal money. The next principal money is $e^x(1+\frac{y}{n})^2$

In this way, we compound our principal money $e^x$ for $n$ intervals (to get from $e^x$ to $e^{x+y}$). After compounding $n$ times, the result is obviously $e^x(1+y/n)^n$. As $n$ tends to infinity, this equals $e^xe^y$.

There's another identity that $(e^a)^x=e^{ax}$.

You can look at the derivative of $(e^a)^x$ to see why that indentity works. The derivative is $a(e^a)^x$. The base $e^a$  just means we're compounding $a$ times faster than the usual base $e$, for $x$ units of time. Rate and time are obviously multiplicative in calculating interest. At $0$, the function equals $1$. At $0+\frac{x}{n}$, it equals $1(1+a\frac{x}{n})$. After compounding $n$ times, it equals ${(1+\frac{ax}{n})^n}$, which is $e^{ax}$ as $n$ tends to infinity.