$ e^{-\mu } \sum_{k=m }^{\infty} \frac{\mu^k}{k!} \le (\frac{\mu}{m})^m e^{m-\mu} $ when $1\le \mu \le m$

Question

for two integers $1 \le \mu \le m$, $$ e^{-\mu } \sum_{k=m }^{\infty} \frac{\mu^k}{k!} \le (\frac{\mu}{m})^m e^{m-\mu}. $$ When I try to prove Chernoff bound, I met this inequality as the last step. But I can't prove. WolframAlpha tells me it is true. Can someone help me prove it?

Answer 1

Divide by $e^{-\mu}$, and also divide by $\mu^m$, you get an equivalent inequality $$\sum_{k=m}^\infty \frac{\mu^{k-m}}{k!}\leqslant \left(\frac{e}m\right)^m.$$ LHS is obviously an increasing function in $\mu$, so it suffices to consider the case $\mu=m$. But then it reads (multiply both sides by $m^m$) as $$ \sum_{k=m}^\infty \frac{m^{k}}{k!}\leqslant e^m $$ which follows from $e^m=\sum_{k=0}^\infty \frac{m^{k}}{k!}$.