TITLE: Showing that the exponential expression $e^x (x-1) + 1$ is positive
QUESTION [11 upvotes]: I'm looking at 
$$ f(x) = e^x (x-1) + 1$$
I'm having the feeling (based on the application where I am using it), that $f(x)$ should be strictly positive for $x > 0$. Indeed, Wolfram Alpha plots it as such, with a global minimum of ($f(0)x=0$).
However, I fail to show this. It is trivial for $x \geq 1$, but what for $x < 1$?

REPLY [14 votes]: For $x\in(0,1)$, the inequality $e^x (x-1)+1 > 0$ is equivalent to:
$$ e^x  < \frac{1}{1-x} \tag{1}$$
or to:
$$ 1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\ldots < 1+x+x^2+x^3+\ldots \tag{2} $$
that is trivial.