$$\begin{align*} \frac{d}{dx} e^x &=\ \lim_{\Delta x\to 0}\frac{e^{x+\Delta x}-e^x}{\Delta x}\\ &=\ \lim_{\Delta x\to 0}\frac{e^x\left(e^{\Delta x} -1 \right)}{\Delta x}\\ &=\ e^x\lim_{\Delta x\to 0}\frac{e^{\Delta x}-1}{\Delta x}\\ &=\ e^x \end{align*}$$

The last step uses the conclusion of # 4522