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I have $Z \sim \operatorname{po}(1)$, and must find $E(e^{Z})$.

My method was as follows. I know that for a function $p$ with support $\{x_i \mid i \in I\}$:

$$E(g(X)) = \sum_{i\in I} g(x_i)\cdot p(x_i) $$

So I plop my transformed variable in, and I get the following:

$$E(e^Z) = \sum_{i=0}^\infty e^i \cdot p(z_i)$$

And the poisson probability mass function:

$$E(e^Z) = \sum_{i=0}^\infty e^i \cdot e^\lambda \cdot \frac{\lambda^i}{i!}$$

And I'm not sure how to proceed from here... I took a sneak peak on wolfram, and saw that this sum converges on $e^{(e-1)\cdot \lambda}$, and since our lambda parameter is $1$, $e^{(e-1)}$. But I'm not sure how can, in hand, reach this result.

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Your basically there with your approach. I think you forgot a minus sign in your probability mass function $p$, i.e. for $k \in \{0,1,\ldots\}$ it should be \begin{align} \Pr[Z = k] = p(k) = e^{-\lambda} \frac{\lambda^k}{k!}. \end{align} Your expectation then becomes \begin{align} \mathbb{E}[\exp(Z)] = \sum_{k=0}^{\infty} e^k e^{-\lambda} \frac{\lambda^k}{k!} = e^{-\lambda} \sum_{k=0}^{\infty} \frac{(e \lambda)^k}{k!}. \end{align} Comparing the last sum with the definition of the exponential function you see that it is equal to $\exp(e\lambda)$. This then leads to \begin{align} \mathbb{E}[\exp(Z)] = e^{(e-1)\lambda} \end{align} and for $\lambda = 1$ you get your result.

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    Ugh. Thanks! It was my power rules that brought me down again. I forgot that e^k * lambda^k can be written as (e*lambda)^k. I think the exponential function series would have jumped out to me quicker if I remembered :) – 2017-01-03