We assume $X \sim Poi(\lambda)$, and want to prove that for every $i<\lambda$, $P(X\leq i)\leq e^{-\lambda}\frac{(e\lambda)^i}{i^i}$. I tried first assuming $i>0$ and following definition:
$P(X\leq i)=\sum_{{k<i}}e^{-\lambda}\frac{\lambda^k}{k!}$ and bounding by $\sum_{{k<i}}e^{-\lambda}\frac{\lambda^i}{k!}$.
So we get $P(X\leq i)=\sum_{{k<i}}e^{-\lambda}\frac{\lambda^k}{k!}\leq e^{-\lambda}\lambda^i\sum_{{k<i}}\frac{1}{k!}$, which can be then bounded by just taking the sum over all the natural numbers, meaning $$P(X\leq i)\leq e^{-\lambda}\lambda^i\sum_{{k<i}}\frac{1}{k!}\leq e^{-\lambda}\lambda^ie$$, but that's the best bound I could find. What am I missing here?
It is not true in general:
Take $\lambda = 2, i =5$ then $P(X \le 5) \approx 0.98344$ and yet $e^{-2}\frac{(2e)^5}{5^5} \approx 0.205675$.
EDIT: I just noticed OP had put $i<\lambda$