\begin{aligned}
(\omega+1)^{\omega+1}
& = (\omega+1)^\omega \cdot (\omega+1) \\
& = \Big(\sup_{n\in\omega} (\omega+1)^n\Big)\cdot(\omega+1) \\
& = \Big(\sup_{n\in\omega} (\omega^n+\omega^{n-1}+\cdots+\omega+1)\Big)\cdot (\omega+1) \\
& = \omega^\omega\cdot(\omega+1) \\
& = \omega^{\omega+1}+\omega^\omega
\end{aligned}
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