\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}