Bound for a sum of poisson probabilities I'd like to understand a bound that has appeared in a paper I'm reading. It has the following expression:

$$\begin{align*}[...] &\leq \sum_{m \geq n }4^mP[Poisson(kt)\geq m]
\\&=e^{-kt}\sum_{m \geq n}4^m \sum_{j \geq m}\frac{(kt)^j}{j!}
\\&\leq e^{-kt}\sum_{m\geq n}e^{kt}\frac{(4kt)^m}{m!}\\
&{\color{red}\leq} e^{4kt}\frac{(4kt)^n}{\left(\frac{n}{2}\right)^{\frac{n}{2}}}
\\&{\color{blue}\leq} e^{-\frac{1}{8}n \log n}\end{align*}$$
if $n \geq (16kt)^8$.

Now, I do understand everything except the two colored inequalities.

For the blue one, I believe the idea is the following: Since $n\geq(16kt)^8\implies 4kt\leq \frac{n^{\frac{1}{8}}}{4}$ and so
$$e^{4kt}\frac{(4kt)^n}{\left(\frac{n}{2}\right)^{\frac{n}{2}}}\leq e^{\frac{n^{1/8}}{4}}\frac{\left(\frac{n^{1/8}}{4}\right)^n}{\left(\frac{n}{2}\right)^{n/2}} = e^{\frac{n^{1/8}}{4}} n^{n/8}n^{-n/2}4^{-n}2^{n/2} = e^{\frac{n^{1/8}}{4}}n^{-\frac{3}{8}n}2^{-\frac{3}{2}n}$$
$$=\exp\left(\frac{n^{1/8}}{4}-\frac{3}{8}n \log n -\frac{3}{2}n \log 2\right)$$
Now, for $n$ big enough $\frac{n^{\frac{1}{8}}}{4}-\frac{3}{2}n \log 2<0$ and so
$\exp\left(\frac{n^{1/8}}{4}-\frac{3}{8}n \log n -\frac{3}{2}n \log 2\right)\leq \exp\left(-\frac{3}{8}n \log n\right)\leq \exp \left(-\frac{1}{8}n \log n\right)$

Now, the red one I have absolutely on idea on how to proceed.

For the red inequality, we have
\begin{equation}
\sum_{m=n}^\infty\frac{(4kt)^m}{m!}=\sum_{m=0}^\infty \frac{(4kt)^{n+m}}{(m+n)!}
\end{equation}
Since $n!\geq(\frac{n}{2})^{\frac{n}{2}}$ and $(m+n)!\geq m!n!$, it follows that the last sum is less than or equal to
\begin{equation}
\frac{(4kt)^n}{(\frac{n}{2})^{\frac{n}{2}}}\sum_{m=0}^\infty\frac{(4kt)^m}{m!}=\frac{(4kt)^n}{(\frac{n}{2})^{\frac{n}{2}}}e^{4kt}.
\end{equation}
