I would like to have an estimate for the series
$$P(t) = \sum\limits\_{k = 0}^\infty (e^{-t}\frac{t^k}{k!})^m,$$
where $e$ is the base of natural logarithm, $k!$ is the factorial of the integer $k$, $t$ represents the time and $t>0$, $m$ is a positive integer and $m>1$ (Obviously, $P(t)=1$ when $m=1$ since $P(t)$ is exactly the cdf of Poisson distribution with associated parameter $t$ in this case.).


I am interested in showing that the answer could look something like 
$$P(t) = O(t^{-\alpha m}),\alpha > 0.$$


It has come to my attention that in the paper “[Rumors in a Network: Who's the Culprit?](http://snap.stanford.edu/nipsgraphs2009/papers/zaman-paper.pdf)” D Shah and T Zaman proved that (Page 24-27)
$$P(t) = O(\frac{1}{\sqrt{t}}),\text{if }m=2.$$


However, I find it difficult for me to extend their method to the cases when $m>2$. 
  
 I also tried to calculate $P(t)$ in Mathematica and it gave me the result like
$$P(t) = \sum\limits\_{k = 0}^\infty (e^{-t}\frac{t^k}{k!})^m = e^{-mt}\text{HypergeometricPFQ}[...,t^m]$$
It seems that the decreasing speed of $P(t)$ is much slower than that of $e^{-mt}$ as $t$ increases. But I could not find the closed-form bound when $m>2$ just like $O(1/\sqrt t)$ when $m=2$. 


Does anyone know of a scale or a bound of $P(t)$ in the literature? Any comments and answers would be highly appreciated. Many thanks!



Estimate the scale of the power series with Poisson pdf/pmf-like terms
Actually this can be solved by standard asymtotic analysis methods (with help from Mathematica):


Expand the factorial in the sum with Sterling. Then do as if $k$ is continuous and find $k=k\_{0}$ where the term in the sum is maximal (the result contains Lambert's $W$ function). Next expand quadratically around $k$ with expansion parameter, say, $\delta$. The quadratic expansion is then re-interpreted as quadratic expansion of a Gaussian function, which one can integrate over $-\infty<\delta<\infty$. The integral converges and one can expand its value (with big help from Mathematica) for large $t$. The result is
$$
P(t)\sim (2\pi t)^{-(m-1)/2}m^{-1/2},
$$
and, indeed, direct tests with Mathematica confirm the result. 


