\begin{aligned}
 \sum_{i\in\omega+1}\prod_{j\in i}(\omega+(i+1)^j)
 & = \sum_{i\in\omega}(\omega^i+\text{ne\v{s}to})+\omega^\omega \\
 & = \sup_{n\in\omega}(\omega^{n-1}+\text{ne\v{s}to})+\omega^\omega \\
 & = \omega^\omega+\omega^\omega \\
 & = \omega^\omega\cdot 2
\end{aligned}