Find expectation $ E[S]$ Let $ N$  be a Poisson random variable $P(N = n) = \frac{\theta^n}{n!}e^{-\theta}$ for n = 0, 1, 2, 3... and $S_n = X_1 + X_2 + ... + X_N$ , where $X_i$ is an independent normal random variables. I need to find $E[S]$ and then prove that $E[t^s] = exp(\theta(e^{\frac{t^2}{2}} - 1))$. 
I think the expression $E[S = s|N = n] = P(S = s)P(S = s|N = n)$ is useful to find out the conditional expectation. And then I can $E[S] = E[S = s|N = n]P(N = n)$ to find out $E[s]$. I have trouble in calculating conditional expectation. Any ideas?
 A: I think you mean $S=X_1+X_2+\ldots+X_N$.  I also assume, from the formula to be proven, that each $X_i$ is a standard normal random variable.  And we are to find $E[t^S]$.
First, we compute 
$$E[t^S|N=n]=E\left[t^{X_1+X_2+\ldots+X_n}\right].$$
By independence of $X_i$ and since the random variables $X_i$ are identically distributed,
$$E[t^S|N=n]=\prod_{i=1}^nE[t^{X_i}]=\Big(E\left[t^{X_1}\right]\Big)^n=\left(\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}t^xe^{-\frac{x^2}{2}}dx\right)^n.$$
Let $z=\ln t$.  Then
$$\int_{-\infty}^\infty t^xe^{-\frac{x^2}{2}}dx=e^{\frac{z^2}{2}}\int_{-\infty}^{\infty}e^{-\frac{(x-z)^2}{2}}dx=e^{\frac{z^2}{2}}\int_{-\infty}^{\infty}e^{-\frac{x^2}{2}}dx=\sqrt{2\pi}e^{\frac{\ln^2t}{2}}.$$
Therefore
$$E[t^S|N=n]=e^{n\frac{\ln^2t}{2}}.$$
This implies
$$E[t^S]=\sum_{n=0}^\infty E[t^S|N=n]\ P[N=n]=\sum_{n=0}^\infty e^{n\frac{\ln^2t}{2}}\frac{\theta^n}{n!}e^{-\theta}.$$
That is
$$E[t^S]=e^{-\theta}\sum_{n=0}^\infty\frac{\left(\theta e^{\frac{\ln^2t}{2}}\right)^n}{n!}=e^{-\theta}e^{\theta e^{\frac{\ln^2t}{2}}}=\exp\Biggl(\theta\left(e^{\frac{\ln^2t}{2}}-1\right)\Biggr).$$
This is not the same as your formula (see my comment under your question to understand why your fomula should be incorrect).
However, if you meant to find the moment-generating function $E\left[e^{tS}\right]$, then yes
$$E\left[e^{tS}\right]=E\left[(e^t)^S\right]=\exp\Biggl(\theta\left(e^{\frac{\ln^2(e^t)}{2}}-1\right)\Biggr)=\exp\Biggl(\theta\left(e^{\frac{t^2}{2}}-1\right)\Biggr).$$
I am not sure why you are trying to find $E[S]$ since $E[S|N=n]=0$ always, so $E[S]=0$.
