Prove that $\lim_{x\to 0}\frac{e^x-1}{x}=1$. I need help understanding this proof:

Prove that $$\lim_{x\to 0}\frac{e^x-1}{x}=1.$$
For $x>0$ and $n\in\Bbb N$: $$1\leq\frac{(1+\frac {x}{n})^n -1}{x}=\frac{1}{n}[(1+\frac{x}{n})^{n-1}+...+1]\leq(1+\frac{x}{n})^{n-1}$$
For $n\rightarrow \infty$ we have
$$1\leq\frac{f(x)-1}{x}\leq f(x)$$.
For $x<0$ by substituting $x$ with $-x$ and dividing by $f(-x)>0$ we get:
$$\frac{1}{f(-x)}\leq \frac{\frac{1}{f(-x)}-1}{x}\leq 1$$
Now for all $x\neq 0$ we have:
$$\min\{1,e^x\}\leq\frac{e^x-1}{x}\leq\max\{1,e^x\}$$.
Using $\lim_{x\to 0} e^x=1$ and the sandwich rule we get: $$\lim_{x\to 0}\frac{e^x-1}{x}=1$$

I know I'm missing out on something important here but how did they get $(1+\frac{x}{n})^n$ from $e^x$ in the very first step?

HINT: $$\lim_{n \to 0} \left(1 + \frac{x}{n}\right)^n = e^x$$
