Remove the first two exponential-series terms

$\begin{aligned}&\sum_{r=2}^{\infty}\frac{x^r}{r!}\\ & =\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots.\end{aligned}$

MF27 gives $e^x=1+x+x^2/2!+x^3/3!+\cdots$. Therefore the required sum is

$e^x-1-x.$

The exponential series converges for every real x.