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.

Let $P_n(x)$ your polynomial. The Taylor-Lagrange formula gives for $x\in \mathbb{R}$ that there exists $c$ such that $\exp(x)=P_n(x)+\frac{x^{2n+1}}{(2n+1)!}\exp(c)$ for some $c$ depending on $x$. Now suppose that there exists $x\in \mathbb{R}$ such that $P_n(x)=0$. Obviously, we have $x<0$. Then $\exp(x)=\frac{x^{2n+1}}{(2n+1)!}\exp(c)$, and $\exp(x-c)=\frac{x^{2n+1}}{(2n+1)!}<0$, a contradiction. Hence $P_n(x)$ is never zero, and as a continuous function, it has a constant sign. As $P_n(0)=1$, we are done. 
