Expectation of Product Poisson Process Let $N(t)$ be a Poisson Process with intensity $\lambda$ letting $Y_i$'s be iid random variables is it true that 
$$E\left(\prod_{i=1}^{N(t)}Y_i\right)=\left(E(Y)\right)^{\lambda t}$$
I know there is a result similar to this when its a sum but wanted to know if this translated over to products. 

I calculate a different identity:
Let $\mu$ be the mean of each of the $Y_i$ random variables.  I define the product to be of the random variables to be $1$ if $N(t)=0$.  See below if you'd like to define it to be $0$.
\begin{eqnarray*}
E\left(\prod_{i=1}^{N(t)}Y_i\right) &=& E\left[E\left(\prod_{i=1}^{N(t)}Y_i| N(t)\right)\right] \\
     &=& \sum_{n=0}^\infty E\left(\prod_{i=1}^{N(t)}Y_i| N(t) = n\right) P(N(t) = n) \\
     &=& \sum_{n=0}^\infty E\left(\prod_{i=1}^{n}Y_i\right) \frac{(\lambda t)^ne^{-\lambda t}}{n!} \\
    &=& \sum_{n=0}^\infty \mu^{n} \frac{(\lambda t)^ne^{-\lambda t}}{n!} \\
    &=& e^{-\lambda t}e^{\mu \lambda t} \\
    &=& e^{(\mu - 1)\lambda t} \\
\end{eqnarray*}
The first line is by the Tower Rule.  If you'd like to choose $0$ for the product $0$ random variables, the same calculation holds, just subtract $e^{-\mu\lambda t}$ from the result.
