Moment-generating function of a Poisson distribution ($M_N(t)=\exp (\lambda (e^t-1))$)

For $N\sim \mathrm{Pois}(\lambda )$ with $\lambda \ge 0$, insert its probability mass function into the expected value:The exponential series shows finiteness for every real $t$. Differentiating this moment-generating function gives mean and variance both $\lambda$. Taking the logarithm gives the cumulant-generating function $\lambda (e^t-1)$, so every positive-order cumulant is also $\lambda$.