Can't figure out a step in the proof of $\cosh^{-1}x=\ln(x+\sqrt{x^2-1}), \forall x\ge1$? I can't figure out the part start from $*$ in the proof from my book:
Set $$y=\cosh^{-1}x,\,x\ge1$$
and note that $$\cosh y=x\quad and\quad y\ge0.$$
... .
$(*)$Since $y$ is nonnegative,
$$e^y=x\pm\sqrt{x^2-1}$$
cannot be less than $1$.
This renders the negative sign impossible.
... .

Note that we have $$\cosh y=\frac{e^y+e^{-y}}{2}=x$$
So we have by multiplying $2e^{y}$ on each side, $$e^{2y}-2xe^{y}+1=0$$
 Setting $e^{y}=t$ and applying the quadratic formula, we have that $$e^{y}=x \pm \sqrt{x^2-1}$$
Note that $\cosh^{-1} x$ is a function that goes from the real numbers which are greater than $1$ to the non-negative reals. So we have that $y \ge 0$. Since $e^{y}$ is an increasing function, $$e^{y}\ge e^{0}=1$$
However,$$x-\sqrt{x^2-1}=\frac{1}{x+\sqrt{x^2-1}} \le \frac{1}{1+\sqrt{1^2-1}}=1$$
So $e^{y} \neq x-\sqrt{x^2-1}$ given that $x \neq 1$. In the case when $x=1$, we have that $x-\sqrt{x^2-1}=x+\sqrt{x^2-1}$. 
Thus, $e^{y}=x+\sqrt{x^2-1}$. 
