Assume that $X_i \sim \text{Poisson}(\lambda^i)$, then we want to find the maximum likelihood estimate (MLE) of $\lambda$ and its asymptotics. I did in the following way, but got stuck here.
Since $\mathbb{P}(X_i=x_i)=e^{\lambda^i}\frac{\lambda^{i x_i}}{x_i!}$. Then the likelihood is $\mathcal L(\lambda;X)=\prod_i^n e^{\lambda^i}\frac{\lambda^{i x_i}}{x_i!}$. And the loglikelihood is $\ell(\lambda;X)=C-\sum_i^n \lambda^i+(\log\lambda)\sum_{i=1}^n ix_i$. By taking derivative w.r.t. $ \lambda$, I got $-\sum_i^n i\lambda^{i-1}+\frac{\sum_{i=1}^n ix_i}{\lambda}=0$, i.e. $\sum_i^n i \lambda^i=\sum_{i=1}^n ix_i$. But I don't know how to proceed to find the MLE of $\lambda$, then.