For any real \(\displaystyle x\), pick \(\displaystyle N \in \mathbb N\) such that \(\displaystyle N > x\) and break the sum into two parts:

\(\displaystyle \sum \limits_{n=0}^\infty {x^n \over n!} = \sum \limits_{n=0}^{N-1} {x^n \over n!} + \sum \limits_{n=N}^\infty {x^n \over n!} < \sum \limits_{n=0}^{N-1} {x^n \over n!} + {x^N \over N!} \sum \limits_{n=0}^\infty {x^n \over N^n} \)

The first term is a finite sum and the summation in the second term is a convergent geometric series.