I have $ X $ ~ $ Pois(\lambda)$, and I need to show that for all $ n > \lambda $ :
$$ Pr(X \ge n ) \le e^{-\lambda} \bigg(\frac{e\lambda}{n}\bigg)^n $$
I sense this is just algebra but I cant get around it.
From Chernoff's bound I can get:
$$ Pr(X \ge n ) \le M_x(t) e^{-tn} $$
And I know that:
$$ M_x(t) = e^{-\lambda}\sum_{k=0}^{\infty}\frac{(\lambda e^t)^k}{k!}$$
So it comes up to showing that :
$$e^{-tn} \sum_{k=0}^{\infty}\frac{(\lambda e^t)^k}{k!} $$
is less than or equal to
$$ \bigg(\frac{e\lambda}{n}\bigg)^n$$
But I cant seem to find the correct calculation, I tried the taylor series of the exponent but I never found my way..
For any $t>0$ $$ Pr(X \ge n ) \le M_x(t) e^{-tn} = e^{\lambda e^t}e^{-\lambda} e^{-tn} \tag{1} $$ Since this ineqiality holds for all $t$, we can take infimum of the r.h.s. over $t$, and the inequality remain true.
You need not to show that r.h.s. in (1) is less than the bound you need. It is sufficient to show that $e^{-\lambda} \left(\frac{e\lambda}{n}\right)^n$ equals to the smallest value of r.h.s. in (1).
Minimal value of $e^{\lambda e^t}e^{-\lambda} e^{-tn}$ is attained at $t=\ln(n/\lambda)>0$. Substitute it to the r.h.s. of (1) and get desired bound.