Ordinals: if $\alpha < \omega^{\beta}$ then $\alpha + \omega^{\beta} = \omega^{\beta}$

Question

I'm trying to proof that if $\alpha < \omega^{\beta}$ then $\alpha + \omega^{\beta} = \omega^{\beta}$, where $\omega$ is the least infinite ordinal.
I started with transfinite induction on $\beta$ by proving it for $\beta = 0$ and $\beta =1$.
But now I'm having problems with the successor and limit cases.
Any hints? Thanks.

Answer 1

Note that if $\beta=\gamma+1$ then $\alpha+\omega^{\gamma+1}=\alpha+\omega^\gamma\cdot\omega$. Now if $\alpha<\omega^\gamma$ we are done; otherwise we can write it as $\omega^\gamma\cdot n+\delta$, such that $\delta<\omega^\gamma$. And so $$\alpha+\omega^{\gamma+1}=\omega^\gamma\cdot n+\delta+\omega^\gamma\cdot\omega=\omega^\gamma\cdot n+(\delta+\omega^\gamma\cdot\omega)=\omega^\gamma\cdot n+\omega^\gamma\cdot\omega=\omega^\gamma(n+\omega)=\omega^{\gamma+1}.$$

The limit case can be solved using the fact that $\omega^\delta=\sup\{\omega^\gamma\mid\gamma<\delta\}$.