Prove for compound Poisson process $\mathbb{E}e^{i\theta X_t}=\mathrm{exp}(-\lambda t\int (1-e^{i\theta x})\mu(dx)),\ \theta\in\mathbb R$. 
Let $N=(N_t)_{t\ge 0}$ be a Poisson process with parameter $\lambda$, and $(\xi,\xi_1,\xi_2,\cdots)$ be an independent sequence of i.i.d. random variables that are independent of $N$, such that the random variable $\xi$ has the distribution of $\mu$. Define the compound Poisson process $X=(X_t)_{t\ge 0}$ by $X_t=\xi_1+\cdots +\xi_{N_t}$. Prove that
  $$\mathbb{E}e^{i\theta X_t}=\mathrm{exp}(-\lambda t\int (1-e^{i\theta x})\mu(dx)),\ \theta\in\mathbb R.$$

I have tried to insert $\mathbb{P}(X_t=k)=\sum_n \mathbb{P}(X_t=k|N_t=n)\mathbb{P}(N_t=n)$ into the left side and get
$$\mathbb{E}e^{i\theta X_t}=\sum_k e^{i\theta k}\sum_n \mathbb{P}(\xi_1+\cdots+\xi_n=k)\cdot \mathbb{P}(N_t=n),$$
interchanging the order of summation yields
$$\mathbb{E}e^{i\theta X_t}=\sum_n \mathbb{P}(N_t=n)\sum_k e^{i\theta k}\cdot \mathbb{P}(\xi_1+\cdots+\xi_n=k).$$
From there I am stuck at calculating the sums.
Thank you for your help!
 A: Using Law of Total Expectations ("Tower Rule"), we have
$$\mathbb{E}\left[ e^{i\theta X_t}\right]= \mathbb{E}\left[\mathbb{E}\left[ e^{i\theta X_t}|N_t\right]\right]$$
Now,
$$\mathbb{E}\left[ e^{i\theta X_t}|N_t=n\right]=\mathbb{E}\left[ exp\left\{i\theta \sum_{i=1}^{n}\xi_i\right\}\Big|N_t=n\right]= \prod_{i=1}^{n}\mathbb{E}\left[ e^{i\theta \xi_i}\right]= \left(\mathbb{E}\left[ e^{i\theta \xi_1}\right]\right)^n$$
where we used the fact that $\xi_i$'s are independent and identically distributed. This computes the inner expectations. 
Now, we need to compute the outer expectations. We have $$ \mathbb{E}\left[\mathbb{E}\left[ e^{i\theta X_t}|N_t\right]\right]=\mathbb{E}\left[\left(\mathbb{E}\left[ e^{i\theta \xi_1}\right]\right)^{N_t}\right]=\sum_{n=0}^{\infty}e^{-\lambda t}\frac{(\lambda t)^n}{n!}\left(\mathbb{E}\left[e^{i\theta \xi_1}\right]\right)^{n} \\
=e^{-\lambda t} \sum_{n=0}^{\infty}\frac{(\lambda t \mathbb{E}\left[e^{i\theta \xi_1}\right])^n}{n!}$$
We recognize that the above sum is simply equal to
$$ \sum_{n=0}^{\infty}\frac{(\lambda t \mathbb{E}\left[e^{i\theta \xi_1}\right])^n}{n!}= e^{\lambda t \mathbb{E}\left[e^{i\theta \xi_1}\right]} = exp\left\{\lambda t\int e^{i\theta x}\mu(dx)\right\}$$ Thus, we obtain
$$ \mathbb{E}\left[ e^{i\theta X_t}\right] =exp\left\{\lambda t\int \left(e^{i\theta x}-1 \right)\mu(dx)\right\} $$
as required.
