Prove that $e^x > 1 + (1 + x)\log(1 + x), x > 0$ using power series expansion. Prove that $e^x > 1 + (1 + x)\log(1 + x), x > 0$ using power series expansion.
I am a bit puzzled by this statement because the power series for $\log(1+x)$ only converges iff $|x|<1$. Is the problem sound? Should it be $1>x>0$?

For $0<x<1$, does the following proof work?
$e^x>1+x+\dfrac{x^2}{2}+\dfrac{x^3}{6}$ and $\log(1+x)<x-\dfrac{x^2}{2}+\dfrac{x^3}{3}$ and so,
$$1+(1+x)\log(1+x)<1+x+\frac{x^2}{2}-\frac{x^3}{6}+\frac{x^4}{3}$$
We observe that $$1+x+\frac{x^2}{2}-\frac{x^3}{6}+\frac{x^4}{3}<1+x+\dfrac{x^2}{2}+\dfrac{x^3}{6}$$
iff
$x^3>x^4$ which is true for $x\in (0,1)$.
