Cantors normal form of x is defined as the following $x = \omega^{a_1} n_1 + \dots + \omega^{a_k} n_k$, Where $x$ is an ordinal and where $\langle a_i \rangle$ is a strictly decreasing finite sequence of ordinals, $\langle n_i \rangle$ is a finite sequence of ordinals and $k\in \Bbb N$.
My problem in the understanding of the cantor normal form is the fact that i don't understand why one can write any finite ordinal in terms of cantors normal form, and why each ordinal has a unique cantor normal form.
First, prove that the map $\alpha\mapsto\omega^\alpha $ is normal, that is, strictly increasing and continuous at limits. Use this to show that for any $\alpha $ there is a least $\beta $ such that $\alpha <\omega^\beta $, and that, if $\alpha\ne0$, then this least $\beta $ is a successor ordinal, say $\beta=\beta_0+1$.
This shows that, for $\alpha\ne0$, there is a unique $\beta_0$ such that $\omega^{\beta_0}\le\alpha <\omega^{\beta_0+1}=\omega^{\beta_0}\cdot\omega $. Conclude from this that there is a unique positive integer $n_0 $ such that $\omega^{\beta_0}\cdot n_0\le \alpha <\omega^{\beta_0}\cdot (n_0+1) $.
Conclude from the above that there is a unique $\gamma <\omega^{\beta_0} $ such that $\alpha=\omega^{\beta_0}\cdot n_0+\gamma $. Now argue inductively, with $\gamma $ in place of $\alpha $.
The argument shows existence of the normal form. Uniqueness follows easily as well: Given two potential representations of $\alpha $, check that they are equal term by term by contradiction, considering the first term from left to right where they disagree.