The function $f(x) = e^x$ is quite peculiar: it is the only function (up to scale) whose derivative is itself. We have \begin{eqnarray*} f'(x) & = & \lim_{h \to 0} \frac{f(x+h)-f(x)}{h} \cr & = & \lim_{h \to 0} \frac{e^{x+h} - e^x}{h} \cr & = & \lim_{h \to 0} \frac{e^xe^h - e^x}{h} \cr & = & e^x \cdot\left(\lim_{h \to 0} \frac{e^h - 1}{h}\right). \end{eqnarray*} The last step was made possible by the fact that $e^x$ doesn't depend on $h$. Notice that $$ \lim_{h \to 0} \frac{e^h - 1}{h}=f'(0), $$ which we have defined to be 1 earlier. This implies that the derivative of $e^x$ is $e^x$!