minimum sufficient statistics problem $X_1\sim \operatorname{Poisson}(\lambda)$, and $X_i\mid X_{i-1}=x_{i-1}\sim \operatorname{Poisson}(\lambda x_{i-1})$ for $i=1,\ldots,n$.
The question asks to find a Minimum Sufficient Statistics and MLE for $\lambda$.
So, I tried to solve it by Lehmann–Scheffe Theorem, I want to find the joint pdf. However, This joint pdf is so giant, and I got stuck on some of the steps, can anyone help a little bit, thanks!

Can get the joint PMF by induction $$ f(x_1,x_2) = f(x_2\mid x_1) f(x_1),\\f(x_1,x_2,x_3) = f(x_3\mid x_2,x_1)f(x_1,x_2) = f(x_3\mid x_2)f(x_2\mid x_1) f(x_1),$$ etc where we used the Markov property.  So the joint PMF for three is $$ f(x_1,x_2,x_3) = \frac{e^{-\lambda x_2}(\lambda x_2)^{x_3}}{x_3!} \frac{e^{-\lambda x_1}(\lambda x_1)^{x_2}}{x_2!}\frac{e^{-\lambda}(\lambda)^{x_1}}{x_1!}.$$
The log-likelihood is then $$ \ell(\lambda)= -\lambda(x_2+x_1+1) + (x_3+x_2+x_1)\log(\lambda) + \text{const}$$ So it looks like $x_3+x_2+x_1$ and $x_2+x_1$ are jointly sufficient (or put more simply $x_3$ and $x_1+x_2$ are). This can be generalized to any $n$.
