Let $h=e^x-1$, then $x\to 0\implies h\to 0$. It follows that

$$\begin{align*} \lim_{x\to 0}\frac{e^x-1}{x}&=\ \lim_{h\to 0}\frac{h}{\ln(1+h)}\\ &=\ \lim_{h\to 0}\frac{1}{\frac{1}{h}\ln(1+h)}\\ &=\ \lim_{h\to 0}\frac{1}{\ln(1+h)^{\frac{1}{h}}} \end{align*}$$

When $h\to 0$, we have $(1+h)^{\frac{1}{h}}=e$. Therefore, the above relation leads to

$$\lim_{x\to 0}\frac{e^x-1}{x}=\lim_{h\to 0}\frac{1}{\ln(1+h)^{\frac{1}{h}}}=\frac{1}{\ln e}=1$$