Compound Poisson Process, probability generating function

Question

If we have a iid random variables $X_i$ with probability generating function $\xi(t) = E[t^{x_i}]$ and $N$ is Poisson with mean $\lambda$ with Probability generating function :
$$ \begin{aligned} \phi(t) = \sum_{i \geq 0} { e^{- \lambda} \lambda ^i \over i!} t^i = e^{-\lambda (1-t)} \end{aligned} $$

Take $Y = \sum_1^NX_i$ what would the probability generating function be in this case?

$$ \begin{aligned} E[t^Y] = E[t^{\sum_1^N X_i}] = E[t^{X_1} ... t^{X_N} ] = E[t^{X_1}] ... E[t^{X_N}] = \xi(t)^N \end{aligned} $$ But this would be assuming that N is not a random variable. I am confused as to how one would take them both into account?

EDIT

So the probability generating function would be: $$ \begin{aligned} E[E[t^{\sum_1^N X_i}|X_i]|N] = E[\xi(t)^N | N] = \sum_{i \geq 0} { e^{- \lambda} \lambda ^i \over i!} \xi(t)^i = e^{-\lambda (1- \xi (t))} \end{aligned} $$ correct?

Answer 1

Based on comment: You've gotten part way there; now take the expectation of your last term with respect to the probability distribution of NN. – jbowman

$$ \begin{aligned} E[E[t^{\sum_1^N X_i}|X_i]|N] = E[\xi(t)^N | N] = \sum_{i \geq 0} { e^{- \lambda} \lambda ^i \over i!} \xi(t)^i = e^{-\lambda (1- \xi (t))} \end{aligned} $$