Upper bound for $e^x$ 
How do I prove that $$e^x \leq x + e^{x^2}$$ for all $x\in\mathbb R$? 

My probability book (Grimmett and Stirzaker) says that it's a simple exercise but I don't see it. For $x\leq 0$, we have
$$e^x = \sum_{k=0}^\infty \frac{x^{2k}}{(2k)!} + x + \sum_{k=1}^\infty \frac{x^{2k+1}}{(2k+1)!} \leq \sum_{k=0}^\infty \frac{x^{2k}}{k!} + x = e^{x^2} + x.$$
How do I show it for $x>0$?

The case $x\ge1$ is trivial. For $0<x<1$, we have
$$
\frac{e^x-e^{x^2}}{x-x^2}\le e^x\le\frac1{1-x}.
$$
The first inequality is true because the LHS, by mean value theorem, is $e^a$ for some $a\in(x^2,x)$. For the second one, note that $f(x):=(1-x)e^x\le1=f(0^+)$ because $f$ is decreasing ($f'(x)=-xe^x<0$) on $(0,1)$.
