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?

Let $e^x-1=y$.
Hence, $y\rightarrow0$ and since $\ln$ is a continuous function, we obtain: 
$$\lim\limits_{x\rightarrow0}\frac{e^x-1}{x}=\lim\limits_{y\rightarrow0}\frac{y}{\ln(1+y)}=\lim\limits_{y\rightarrow0}\frac{1}{\ln(1+y)^{\frac{1}{y}}}=1$$
