Let $f: \mathbb{R} \rightarrow \mathbb{R}$ given by $f(x)=e^{\theta x}$, where $\theta \in \mathbb{R}$ and suppose that $X ~ Po(\lambda)$ for some $\lambda >0$. Show that $E[f(X)]=exp(\lambda (e^\theta -1))$.
By theorem, $E[f(x)]=\sum_{x=1}^{\infty}{f(x)P(X=x)}=\sum_{x=1}^{\infty}{e^{\theta x}(\frac{\lambda^xe^{-\lambda}}{x!}})=e^{\theta-\lambda}\sum_{x=1}^{\infty}\frac{e^x\lambda^x}{x!}=e^{\theta-\lambda}(\sum_{x=1}^{\infty}e^x)(\sum_{x=1}^{\infty}\frac{\lambda^x}{x!})=e^{\theta-\lambda}(e^\lambda)(\sum_{x=1}^{\infty}e^x)=e^{\theta}(\sum_{x=1}^{\infty}e^x)$
First question, is my calculation correct? Second question, what is the infinite sum of $e^x$
The sum should start at $0$, and the third equality you have is wrong. Recall that $ e^ae^b = e^{a+b} $ (not $e^ae^b = e^{ab}$); and that $e^{ab} = (e^a)^b$. Using this, you have $$ \mathbb{E}[f(x)]=\sum_{x=0}^{\infty}{f(k)\mathbb{P}\{ X=k \}} = e^{-\lambda}\sum_{x=0}^{\infty}{e^{\theta k} \frac{\lambda^k}{k!}} = e^{-\lambda}\sum_{x=0}^{\infty}{\frac{(e^\theta\lambda)^k}{k!}} = e^{-\lambda}e^{e^\theta\lambda} = e^{(e^\theta-1)\lambda} $$ proving the result. (The last steps use the fact that $\sum_{x=0}^{\infty}{\frac{x^k}{k!}} = e^x$.)