How to show $1 +x + x^2/2! + \dots+ x^{2n}/(2n)!$ is positive for $x\in\Bbb{R}$? 
How to show $1 + x + \frac{x^2}{2!} + \dots+ \frac{x^{2n}}{(2n)!}$ is positive for $x\in\Bbb{R}$?

I realize that it's a part of the Taylor Series expansion of $e^x$ but can't proceed with this knowledge? Also, I can't figure out the significance of $2n$ being the highest power.

If we use Taylor's theorem with an integral remainder we get:
$$ 1+x+\frac{x}{2}+\ldots+\frac{x^{2n}}{(2n)!} = e^x-\frac{1}{(2n)!}\int_{0}^{x}t^{2n}e^{x-t}\,dt \tag{1}$$
hence in order to prove the non-negativity of the LHS it is enough to show that:
$$ \int_{0}^{x} t^{2n}e^{-t}\,dt \leq (2n)! \tag{2}$$
holds for any $x\in\mathbb{R}$. $(2)$ is trivial if $x\leq 0$, and if we assume $x>0$, since the integrand function is non-negative, we get:
$$ \int_{0}^{x}t^{2n}e^{-t}\,dt \leq \int_{0}^{+\infty}t^{2n}e^{-t}\,dt = \Gamma(2n+1)=(2n)!,\tag{3}$$
proving our claim.
