Proof of exponentiation rules with rational exponents It is clear for me how to proof the following rules:
$ a^x\cdot a^y=a^{x+y}$ 
$(a^x)^y=a^{xy}$
for $a\in \mathbb{R}$ and $x,y \in \mathbb{Z}$. 
But i don't understand the step for concluding the same rules for $x,y \in \mathbb{Q}$.
I know from school, that these rules are applicable also for $x,y\in \mathbb{R}$ as well. How can one show this?

We can define the natural logarithm function for positive real numbers by $$\ln(x)=\int_1^x\frac{dt}{t}.$$
This function is one-to-one and has all real numbers in its range, so its inverse can be defined for all real numbers, we we call it $e^x$, or $\exp(x)$. From the properties of logarithms, we deduce that $\exp(x+y)=\exp(x)\exp(y)$.  Once we have this function defined, we can define exponentiation as:
$$a^x=\exp(x\ln a).$$
From this definition, the exponent rules follow:
$$\begin{align}
a^xa^y = \exp(x\ln a)\exp(y\ln a) &= \exp(x\ln a+y\ln a)\\
&=\exp((x+y)\ln a)\\
&=a^{x+y}
\end{align}$$
$$\begin{align}
(a^x)^y = \exp(y\ln a^x) &= \exp(y\ln(\exp(x\ln a)))\\
&=\exp(y(x\ln a))\\
&= \exp(xy\ln a)\\
&=a^{xy}
\end{align}$$
Note that all of this applies to $a>0$ in $\Bbb R$. If $a$ is negative, these rules don't necessarily apply for all $x,y$.
