Define the exponential function in terms of the natural logarithm and show that the derivitave of e^x is e^x
Define $\displaystyle \ln x = \int_1^x \frac{d\mu}{\mu}$, for $\displaystyle x>0$. This function is differenciable by the fundamental theorem for any $\displaystyle x>0$ and $\displaystyle \ln x = \frac{1}{x}$. Since the derivative is positive the function is increasing, so $\displaystyle \ln x$ is a one-to-one function on $\displaystyle \mathbb{R}^+$. Let $\displaystyle \exp (x)$ be its inverse function on [tex]D[tex] (where $\displaystyle D$ is the range of $\displaystyle \ln x$, it happens to be that $\displaystyle D=\mathbb{R}$ but it does not matter here). Thus, $\displaystyle \ln (\exp (x)) = x$. Using the chain rule we get $\displaystyle \left( \exp x \right) ' \frac{1}{\exp x} = 1 \implies \left( \exp x \right) ' = \exp x$.
We can go further and define some additional properties. We know that there is a unique number $\displaystyle e$ such that $\displaystyle \ln e = 1$ by intermediate value theorem. Let $\displaystyle q$ be any rational number. We can show that $\displaystyle \exp ( q ) = e^q$ where $\displaystyle e^q$ is ordinary exponentiation of rational exponents (try provong this). Thus, it is reasonable to define $\displaystyle e^r = \exp (r)$ for any real number $\displaystyle r$.