Ordinal arithmetic $(2\cdot (\omega+1))\cdot \omega$ and $\gamma = \bigcup \{\omega, \omega^\omega, \omega^{\omega^\omega}\}$ Cantor Normalform

Question

Ordinal arithmetic $(2\cdot (\omega+1))\cdot \omega$ and $\gamma = \bigcup \{\omega, \omega^\omega, \omega^{\omega^\omega}, ...\}$ Cantor Normalform

I understood that $(2\cdot (\omega+1))\cdot \omega=(\omega+1)\cdot \omega = \omega\cdot \omega$ for the first one. Is this correct?

I know that the epsilon nought is $\varepsilon _{0}=\omega ^{\omega ^{\omega ^{\cdot ^{\cdot ^{\cdot }}}}}=\sup\{\omega ,\omega ^{\omega },\omega ^{\omega ^{\omega }},\omega ^{\omega ^{\omega ^{\omega }}},\dots \}$, so $\gamma$ should equal $\varepsilon _{0}$.

Is the attempt correct like this? If so, is $\varepsilon _{0}$ in Cantor Normalform? How to proceed?

Answer 1

$2\cdot(\omega+1)=2\cdot\omega+2=\omega+2$, not $\omega+1$, but $(\omega+2)\cdot\omega=\omega\cdot\omega$, so you ended up in the right place. The Cantor normal form of $\epsilon_0$ is $\omega^{\epsilon_0}$.