Let $X$ a positive random variable with power law distribution with exponent $\tau$, meaning that there exists $c>0$ such that its CDF has the form \begin{equation} 1-F(x)=\mathbb{P}(X\ge x)\sim cx^{-(\tau-1)}. \end{equation}
Define the random variable $Y$ defined on $\mathbb{N}^+$to have mixed Poisson distribution with mixing distribution F \begin{equation} \mathbb{P}(Y=k)=\mathbb{E}(e^{-X}\frac{X^k}{k!}). \end{equation} How can one prove that $Y$ has power law distribution with the same exponent $\tau$?
I get stuck when substituting in the integral, since i get \begin{equation} \mathbb{P}(Y\ge x)\approx\sum_{k\ge x}\frac{1}{k!}\int_{0}^{\infty}e^{-y}{y^k}cy^{-\tau}dy \end{equation} and I don't see how it should follow a power law