\begin{aligned}
 (\omega+1)\cdot\omega^{i+2}
 & = (\omega+1)\cdot\omega\cdot\omega^{i+1} \\
 & = \Big(\sup_{n\in\omega}(\omega+1)\cdot n\Big)\cdot\omega^{i+1} \\
 & = \Big(\sup_{n\in\omega}(\omega\cdot n+1)\Big)\cdot\omega^{i+1} \\
 & = \omega^2\cdot\omega^{i+1} \\
 & = \omega^{i+3}
\end{aligned}