Prove that $e^x>x(x+1)$ 
Let $x>0$,prove that $$e^x>x(x+1)$$or
  $$x>\ln{x}+\ln{(x+1)}$$

we can use this Taylor some  first four term,
$$e^x>1+x+\dfrac{1}{2}x^2+\dfrac{1}{6}x^3$$ prove it
But this inequality it seem very nice,maybe there exsit simple methods or Amazing way?

We need to prove$$e^x>1+x+\dfrac{1}{2}x^2+\dfrac{1}{6}x^3\ge x^2+x$$which is$$x^3+3x^2+6x+6\ge 6x^2+6x$$which is$$x^3-3x^2+6=(x-2)^2(x+1)+2>0$$which is obvious for positive $x$.
