Let $F(x)$ be some distribution. It is implicitly defined through
$$a = \sum_{k=2}^\infty (1-F(x))^{k-2}\frac{\lambda^k\exp(\lambda)}{k!} $$
where the right-hand side is basically $(1-F(x))^{k-2} \times Poisson(k, \lambda)$ For the much easier case where the exponents are aligned $\sum_{k=2}^\infty (1-F(x))^{k}\frac{\lambda^k\exp(\lambda)}{k!}$ I can easily solve for $F(x)$ by application of a Taylor expansion. Here, I'm not really sure what to do.
I tried
$$a =(1-F(x))^{-2}\exp(\lambda) \sum_{k=2}^\infty (1-F(x))^k\frac{\lambda^k}{k!} \\ a = (1-F(x))^{-2}\exp(\lambda) \left[ \exp(\lambda(1-F(x)) - 1 - \lambda(1-F(x)) \right] $$
But now I wouldn't know how to solve for $F(x)$. Is there somewhere I can go with this expression? Or is there a different approach that yields a different representation of $F(x)$?