Prove that $0+\beta=\beta$ where $\beta$ is an ordinal.
My attempt was using transfinite induction, and is as follows
We do the induction on $\beta$ so when $\beta=0$ we have that $0+0=0=\beta$ so the case when $\beta=0$ holds.
Now when $\beta$ is a successor ordinal we have that $\beta=S(\gamma)$ for some ordinal $\gamma$ so $0+\beta=S(0+\gamma)=S(\gamma) =\beta$ so the case when $\beta$ is a successor also holds by the definition.
Now when $\beta$ is a limit ordinal specifically $\beta>1$ so we have that $0+\beta=\sup_{\gamma<\beta}(0+\gamma)$ =$\sup_{\gamma<\beta}\gamma$ =$\beta$ hence the case when $\beta$ is a limit ordinal holds also thus by the principle of transfinite induction it holds that $0+\beta=\beta$. $\forall \beta $ ordinals .
I'm unsure if i've understood the principle of transfinite induction correctly, so if someone could clarify if my proof is correct that would be great.
Note: My definition of ordinal addition is that of
$\alpha+\beta =
\begin{cases}
\text{$\alpha$,} &\quad\text{if $\beta=0$}, \\
\text{$S(\alpha+\gamma)$,} &\quad\text{if $\beta=S(\gamma)$} \\
\text{$\sup_{\gamma<\beta}(\alpha+\gamma)$} &\quad\text{if $\beta$ is a limit ordinal}\\
\end{cases}$
Where $S(\gamma)$ is the successor of $\gamma$