Given $N \sim \mathrm{Poisson}(\lambda A)$, what will be $(1 - \frac{c}{A})^{N-1}$, where $c$ is a constant, when $A \to \infty$?
Will it converge into a distribution or a constant? And what is it?
Here is a try:
From @zhoraster, $N$ is distributed as $P_1 + \dots + P_A$, where $P_k$ are iid $\mathrm{Poisson}(\lambda)$.
By the law of large numbers, $\frac{N}{A} \to \lambda$, $n\to\infty$, in probability.
$$ (1 - \frac{c}{A})^{N-1} = \left((1 - \frac{c}{A})^{\frac{A}{c}} \right)^\frac{c(N-1)}{A} = \exp\left(\frac{c(N-1)}{A} \log\left((1 - \frac{c}{A})^{\frac{A}{c}} \right)\right) $$
Since we have $\lim_\limits{A \to \infty} (1 - \frac{c}{A})^{\frac{A}{c}} = e^{-1}$ and $\frac{N}{A} \to \lambda$ as $A \to \infty$. We can obtain $$\lim_{A \to \infty}\frac{c(N-1)}{A} = \lambda c$$ And $$ \lim_{A \to \infty} \log\left((1 - \frac{c}{A})^{\frac{A}{c}} \right) = -1 $$ Thus, $$ \lim_{A \to \infty} \left(\frac{c(N-1)}{A} \log\left((1 - \frac{c}{A})^{\frac{A}{c}}\right)\right) = -\lambda c $$ $$ \lim_{A \to \infty}(1 - \frac{c}{A})^{N-1} = e^{-\lambda c} $$
However, I am not sure if steps are rigorous enough because $A$ can be any positive real number but not necessary a positive integer. Any correction and suggestion is welcome. Thanks!
Hint: $N$ is distributed as $P_1 + \dots + P_A$, where $P_k$ are iid $\mathrm{Poisson}(\lambda)$. So by the law of large numbers, $N/A \to \lambda$, $n\to\infty$, in probability.