Showing that the exponential expression $e^x (x-1) + 1$ is positive

$f'(x)=x \, e^x > 0$ for $x>0$, so $f$ is strictly increasing on $[0,\infty)$.

Since $$ e^x\ge 1+x $$ and thus also $$ e^{-x}\ge 1-x $$ one gets $$ f(x)=e^x·(e^{-x}-(1-x))\ge 0. $$

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.