Prove every nonzero ordinal is either successor or limit

Question

I am trying to prove the statement in the title.

Here is my partial proof:

Let $n$ be a nonzero ordinal. It is obvious that $\cup_{\beta<n}\beta \subseteq n$.

If $\cup_{\beta<n}\beta = n$, then $n$ is a limit ordinal

Otherwise, $\cup_{\beta<n}\beta \subsetneq n$. Then $\exists\,\alpha\in n\;(\alpha\notin\cup_{\beta<n}\beta)$. It follows that $\forall\beta<n\;(\alpha\ge\beta)$

The only thing left to prove is that $n = \alpha\cup\{\alpha\}$, which I can't figure out.

Answer 1

HINT: Show that $\alpha$ is the only ordinal in $n$ which is not an element of the union $\bigcup n$.