Integrating a Poisson Process with respect to itself

Question

I am just learning about Poisson Processes and I feel somewhat comfortable with the basic concepts, but I am a little stuck with the following problem:

Let $N(t)$ be a Poisson process with intensity $\lambda$.

Find $\int_0^t N(s)dN(s) $.

What I have so far is

$\int_0^t N(s)dN(s) = \sum_{0 < s \le t} N(s)\Delta N(s) = \sum_{0 < s \le t} N(s)$

But I am not sure whether or not this is correct. If it is correct, is there any further simplification that is possible?

Answer 1

$$\sum_{0 < s \le t} N(s)\Delta N(s)=\sum_{n=1}^{N(t)}n=\frac{N(t)(N(t)+1)}2\qquad \sum_{0 < s \le t} N(s^-)\Delta N(s)=\sum_{n=1}^{N(t)}n-1=\frac{N(t)(N(t)-1)}2$$