Prove that $\frac{e^x-1}{x}$ is bijective I am trying to prove that this function is bijective, but I don't know how to do it.

$$f:x \mapsto \frac{e^x-1}{x}$$

For that, I try to use the fact that this function has necessarily a reciprocal if it is bijective. 
So, I think I need to prove that :
$$f(x)=y \Leftrightarrow x = f^{-1}(y)$$ 
If I understand correctly, this means that : 
$$\frac{e^x-1}{x} = y \Leftrightarrow x = \frac{y}{e^y-1}$$
But my problem is that I don't know how to do this, and I don't know if there is a more efficient way to solve the problem. 
Can anyone help me solve this problem?
 A: $\newcommand{\R}{\mathbb{R}}$
Let us add some more structure. Consider the function $f: \R\setminus\{0\} \to \R_{+}\setminus \{1\}$ defined by
$$
f(x) = \frac{e^x -1}{x}
$$
and define its extension $g:\R \to \R_{+}$ where $g(0) = 1$. We can easily show that $g$ is continuous at $x = 0$ by L'Hopital's rule.
Now consider the derivative of $g$,
$$
g'(x) = \frac{e^x(x-1) +1}{x^2}
$$
We have
$$
\lim_{x \to 0} g'(x) = \frac{1}{2}
$$
Hence from Hermis14, $g'(0) = 1/2$.
Furthurmore, $g'(x)$ is always positive and $g$ is surjective, which means $g$ is bijective.
Therefore, $f$ is bijective.
A: It's useful to notice that
$$\frac{e^x-1}{x} = \int_0^1 e^{x t} d t$$
so the function $f$ and all its derivatives are $>0$, with $\lim_{x\to -\infty} \frac{e^x-1}{x} = 0$, $\lim_{x\to \infty} \frac{e^x-1}{x} = \infty$.
Obs: The reciprocal $g(x) = \frac{1}{f(x)} = \frac{x}{e^x-1}$ is the generating function of the Bernoulli numbers
A: Set $f(x)=\frac{e^x-1}{x}.$ Then, the claim follows immediately from the following easily verified facts:
$f$ is continuous on $\mathbb R\setminus \{0\},\ \underset {x\to \infty}\lim f(x)=\infty,\ \underset {x\to -\infty}\lim f(x)=0\ $ and $f'(x)=\dfrac{\left(x-1\right)e^x+1}{x^2}>0$ on $\mathbb R\setminus \{0\}.$
