Consider the function $u:x\mapsto\mathrm e^{-x}-1+x$. Its derivative is u':x\mapsto-\mathrm e^{-x}+1. Since $\mathrm e^0=1$ and the exponential is an increasing function, u'(x)\lt u'(0)=0 if $x\lt0$ and u'(x)\gt u'(0)=0 if $x\gt0$. Thus, for every $x$, $u(x)\geqslant u(0)=0$.
This proves that $\sum\limits_{k=2}^{+\infty}\frac{(-x)^k}{k!}\geqslant0$ for every real number $x$, and that $\sum\limits_{k=2}^{+\infty}\frac{(-x)^k}{k!}\gt0$ for every $x\ne0$.
Edit: In the (somewhat odd) situation of not knowing that the sum of the series $s(x)=\sum\limits_{k=2}^{+\infty}\frac{(-x)^k}{k!}$ is $\mathrm e^{-x}-1+x$ but being allowed to differentiate $s(x)$ term by term, one can proceed as follows. First, s'(x)=\sum\limits_{k=2}^{+\infty}(-1)^kk\frac{x^{k-1}}{k!}=\sum\limits_{k=1}^{+\infty}(-1)^{k+1}\frac{x^{k}}{k!}=x-s(x). Hence the function $t:x\mapsto s(x)\mathrm e^x$ is such that t'(x)=(s'(x)+s(x))\mathrm e^x=x\mathrm e^x. In particular, t'(x)\lt0 if $x\lt0$ and t'(x)\gt0 if $x\gt0$. Since $t(0)=0$ this yields $t(x)\gt0$ for every $x$, and finally, $s(x)=\mathrm e^{-x}t(x)\gt0$ for every $x$.