Expectation and Poisson process 
Let {$N(t),t\geq0$} be a Poisson process with rate $\lambda$.
  Calculate $E[N(t).N(t+s)]$

I know that $N(t)\sim Poisson(\lambda t)$ and $N(t+s)\sim Poisson(\lambda(t+s))$
I can assume that $N(t)$ and $N(t+s)$ are independent?
I do not see another way to do it, unless there is some trick.
EDIT: $$E[N(t).N(t+s)]=E[N(t)(N(t+s)-N(t))+N(t)N(t)]$$
$$=E[N(t)(N(t+s)-N(t))]+E[N(t)^2]$$
since $N(t)$ and $N(t+s)-N(t)$ are independent
$$E[N(t)(N(t+s)-N(t)]=E[N(t)].E[N(s)]=\lambda^2ts$$
where $\big(N(t+s)-N(t)\big)\sim Poisson(\lambda s)$
If $N(t)\sim Poisson(\lambda t)$ then $E[N(t)^2]=\lambda t+\lambda^2t^2$ then
$$E[N(t).N(t+s)]=\lambda^2ts+\lambda^2t^2+\lambda t=\lambda t(\lambda ts+\lambda t+1)$$

$N(t)$ and $N(t+s)$ are not independent. On the other hand, $N(t)$ and $N(t+s)-N(t)$ are independent. Moreover, $N(t)N(t+s) = N(t)[N(t+s)-N(t)] + N(t)N(t)$. Now use linearity of expectation and independence. Can you finish it from here?
