Joint expectation in Poisson process

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Let $\{N(t);t ≥ 0\}$ be a Poisson process with intensity $λ > 0$.

We must compute $E(N(1)^2 \cdot N(2))$ without any other information.

Because it is a Poisson process then it has independent increments, but can I use this fact to split this expectation and simplify it as follows : $$E(N(1)^2 \cdot N(2)) = E(N(1)^2) \cdot E(N(2))?$$

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$N(1)$ and $N(2)$ are not independent, so your splitting is invalid. Using independence of $N(1)$ and $N(2)-N(1)$ we get $$E(N(1)^{2}N(2))$$ $$=E(N(1)^{2} ((N(2)-N(1))+E(N(1)^{3})$$ $$=E(N(1)^{2})E(N(2)-N(1))+E(N(1)^{3}).$$

Can you finish?

[$N(1)\sim Poiss (\lambda)$ and $N(2)-N(1)\sim Poiss (\lambda)$].