Please Help me. (Poisson Process: Customer arrival time). 
Customers arrive at a facility according to a Poisson process $N(t)$ of
  rate $\lambda$, with arrival times $S_1$, $S_2$, ...... Each customer pays \$1 on
  arrival. At time $t$, the discounted value of the total sum collected so
  far, discounted back to time zero, is $\sum_{i=1}^{N(t)}e^{- \beta
 S_i}$ , where  $\beta> 0$ is the discount rate. Show that the discounted
  value at time $t$ has expectation $\lambda \beta^{-1}(1-\exp(-\beta
 t))$.

My Attempt:
The sum $\sum_{i=1}^{N(t)}e^{- \beta S_i}=e^{- \beta S_1}+e^{- \beta S_2}+....+e^{- \beta S_{N(t)}}$
If want to take the expected value, then we can just do:
$n*E[e^{- \beta S_i}]$, which is easy to calculate. However, since $N(t) $ itself is also a random variable, I don't know how to calculate $n$.
What about condition on $N(t)=k$?
 A: We apply the total expectation theorem as follows:
$$
\mathbf{E} \sum_{i=1}^{N(t)} e^{-\beta S_i}=\sum_{k=0}^{\infty} \mathbf{E} \left[\sum_{i=1}^{N(t)} e^{-\beta S_i}| N(t)=k\right]e^{-\lambda t}\frac{(\lambda t)^k}{k!}.$$
Then we need to compute 
$$
\mathbf{E} \left[\sum_{i=1}^{N(t)} e^{-\beta S_i}| N(t)=k\right] = \mathbf{E} \left[\sum_{i=1}^{k} e^{-\beta S_i}| N(t)=k\right].$$
Given that $N(t)=k$, the arrival times $S_1, \cdots, S_k$ are chosen randomly in the interval $[0,t]$ to satisfy the only restriction $S_1\leq \cdots \leq S_k$. In fact, the $k$-tuples $(S_1, \cdots , S_k)$ are uniformly distributed on $k$-dimentional polygon 
$$
\{(x_1, \cdots , x_k) | 0\leq x_1 \leq \cdots \leq x_k \leq t \},$$
which has volume $t^k/k!$. 
We now fix $1\leq i\leq k$, and compute 
$$
\mathbf{E}\left[e^{-\beta S_i} | N(t)=k\right].$$
This can be found from an integral:
$$
\frac{k!}{t^k}\int_0^t \frac{u^{i-1}}{(i-1)!} e^{-\beta u} \frac{(t-u)^{k-i}}{(k-i)!}du.$$
Summing over $1\leq i\leq k$, we obtain by binomial theorem that
$$
\begin{align}
\mathbf{E} \left[\sum_{i=1}^{k} e^{-\beta S_i}| N(t)=k\right]&=\sum_{i=1}^k \mathbf{E}\left[e^{-\beta S_i} | N(t)=k\right]\\
&=\sum_{i=1}^k \frac{k!}{t^k}\int_0^t \frac{u^{i-1}}{(i-1)!} e^{-\beta u} \frac{(t-u)^{k-i}}{(k-i)!}du\\
&=\frac{k}{t^k}\int_0^t \sum_{i=1}^k \frac{(k-1)!}{(i-1)!(k-i)!} u^{i-1}(t-u)^{k-i} e^{-\beta u}du\\
&=\frac{k}{t^k}\int_0^t t^{k-1} e^{-\beta u} du
=\frac k {t\beta} (1-e^{-\beta t}).
\end{align}
$$
Putting this back into the sum over $k$, the sum becomes 
$$
\frac1{t\beta}(1-e^{-\beta t}) \mathbf{E}N(t).$$
Since $N(t)$ is Poisson distribution of parameter $\lambda t$, we have $\mathbf{E}N(t)=\lambda t$, thus we conclude that 
$$
\mathbf{E} \sum_{i=1}^{N(t)} e^{-\beta S_i}=\frac{\lambda}{\beta}(1-e^{-\beta t}).$$
