Prove that summation is less than 1 Prove that $m! \times \sum_{n=m+1}^\infty \frac{1}{n!} < 1$
I started proving this by simplifying it to:
$= m! \times \left(\frac{1}{(m+1)!} + \frac{1}{(m+2)!} + ...\right)$
$= \left(\frac{1}{(m+1)} + \frac{1}{(m+2)(m+1)} + ...\right)$
I'm not too sure how to show that the last step sum is less than 1. 

If $m\ge 1$
$m!(\frac{1}{(m+1)!}+\frac{1}{(m+2)!}+\frac{1}{(m+3)!}+...)=\frac{1}{m+1}+\frac{1}{(m+2)(m+1)}+\frac{1}{(m+3)(m+2)(m+1)}+...\le \frac{1}{2}+\frac{1}{2^2}+...=1$
