Showing that the exponential expression $e^x (x-1) + 1$ is positive
Solution 1:
$f'(x)=x \, e^x > 0$ for $x>0$, so $f$ is strictly increasing on $[0,\infty)$.
Solution 2:
Since $$ e^x\ge 1+x $$ and thus also $$ e^{-x}\ge 1-x $$ one gets $$ f(x)=e^x·(e^{-x}-(1-x))\ge 0. $$