Answer
$R=\infty$ ; interval of convergence is $(-\infty, \infty)$
Work Step by Step
Let $a_{n}=\frac {x^{n}}{n!}$, then
$\lim\limits_{n \to \infty}|\frac{a_{n+1}}{a_{n}}|=\lim\limits_{n \to \infty}|\dfrac{\frac {x^{n+1}}{(n)!}}{\frac {x^{n}}{n!}}|$
$=0\lt 1$
Hence, $R=\infty$ ; interval of convergence is $(-\infty, \infty)$