Use the ratio test to prove that
converges,at least for x>0.
Definition of ratio test is if $\displaystyle \lim_{n\to\infty}\frac{a_{n+1}}{a_n}<1$ it converges....so $\displaystyle a_{n+1}=\frac{x^{n+1}}{(n+1)!}=\frac{x\cdot{x^{n}} }{(n+1)n!}$ and divding by a_n is the same as multiplying by its reciporcal which would be $\displaystyle \frac{n!}{x^{n}}$