Show that $\exp(x)-1=\mathcal{O}(x)$ for $x\to 0$

Find a function $g(x)$ that is as simple as possible s.t. $\exp(x)-1=\mathcal{O}(g(x))$ for $x\to 0$.

Claim. Such a possible function is $g(x)=x$.

Proof. Using the definition of the class $\mathcal{O}$ via limits and the limit of the difference quotient we get the following

$$\lim_{x\to 0}\frac{\exp(x)-1}{x} = \lim_{x\to 0}\frac{\exp(0 + x)-\exp(0)}{x} = (\exp(0))' \overset{(*)}{=} \exp(0) = 1<\infty.$$

Hence $\exp(x)-1=\mathcal{O}(x).$ $\tag*{$\blacksquare$}$

I was curious now as the evaluation of $\exp(0)$ to $1$ at $(*)$ and deriving it afterwards would yield $0$ instead of $1$ even though both values still would be less than infinity. I was wondering whether there are just multiple alternatives or the actual evaluation would be just wrong in this particular case. What's the actual reasoning in this case?

You can just apply L'Hopital's rule, since the limit of of the form $\frac{0}{0}$: $$\lim_{x \to 0}{\frac{e^x-1}{x}}=\lim_{x \to 0}{\frac{e^x}{1}=1.}$$
EDIT: In your answer, you always differentiate first before you evaluate the function at a point, and so obtain $\displaystyle \left.e^x\right|_{x=0}=e^0=1$ rather than $(e^0)'=1'=0.$