Cantor normal form: every nonzero ordinal is $\omega^{\beta_0}\cdot c_0 + \cdots + \omega^{\beta_{k-1}}\cdot c_{k-1}$ with $\beta_0 > \cdots > \beta_{k-1}$ and each $c_i$ a nonzero natural number, in exactly one way

Preliminaries on $\omega$ and on powers of $\omega$: for $n, m \in \omega$ one has $n + m \in \omega$, by induction on $m$ over the ordinals in $\omega$, since $n + 0 = n$, since $n + m^{+} = (n + m)^{+} \in \omega$ as $\omega$ is closed under successor, and since no ordinal in $\omega$ is a limit by [L7]; and $\omega^{\delta^{+}} = \omega^{\delta} \cdot \omega = \sup\{\omega^{\delta} \cdot n : n \in \omega\}$ is a limit ordinal, with $\omega^{\delta} \cdot n < \omega^{\delta^{+}}$ for every $n \in \omega$, by [L1], [L5] and claims (d) and (g) of [L3].

Additive indecomposability: for every ordinal $\beta$ and every $\mu < \omega^{\beta}$ one has $\mu + \omega^{\beta} = \omega^{\beta}$. By induction on $\beta$. At $\beta = 0$, $\omega^{0} = 1$ forces $\mu = 0$ and $0 + 1 = 1$. At $\beta = \delta^{+}$: $\mu < \omega^{\delta^{+}} = \sup\{\omega^{\delta} n : n \in \omega\}$ gives $n \in \omega$ with $\mu < \omega^{\delta} \cdot n$, and claim (f) of [L3] applied to the nonempty $D = \{\omega^{\delta} \cdot m : m \in \omega\} \subseteq \omega^{\delta^{+}}$ gives $\mu + \omega^{\delta^{+}} = \sup\{\mu + \omega^{\delta} m : m \in \omega\}$, where each $\mu + \omega^{\delta} m \le \omega^{\delta} n + \omega^{\delta} m = \omega^{\delta}(n + m) < \omega^{\delta^{+}}$ by [L4], step 1.1 and claim (d) of [L3]; so $\mu + \omega^{\delta^{+}} \le \omega^{\delta^{+}}$, and the reverse inequality is claim (c) of [L3]. At $\beta = \lambda$ a limit: $\mu < \omega^{\lambda} = \sup\{\omega^{\xi} : \xi \in \lambda\}$ gives $\xi_0 \in \lambda$ with $\mu < \omega^{\xi_0}$, and $D = \{\omega^{\xi} : \xi \in \lambda \text{ and } \xi_0 \le \xi\}$ is nonempty, contained in $\omega^{\lambda}$ and has supremum $\omega^{\lambda}$, because any $\eta < \omega^{\lambda}$ satisfies $\eta < \omega^{\xi}$ for some $\xi \in \lambda$ and $\xi$ may be replaced by the larger of $\xi$ and $\xi_0$; so claim (f) of [L3] gives $\mu + \omega^{\lambda} = \sup\{\mu + \omega^{\xi} : \xi_0 \le \xi \in \lambda\} = \sup\{\omega^{\xi} : \xi_0 \le \xi \in \lambda\} = \omega^{\lambda}$, using the claim at each such $\xi$, legitimate since $\mu < \omega^{\xi_0} \le \omega^{\xi}$.

The leading exponent exists: for $\alpha > 0$ the set $B = \{\beta \in \alpha^{+} : \omega^{\beta} \le \alpha\}$ contains $0$, because $\omega^{0} = 1 \le \alpha$, and it contains every $\beta$ with $\omega^{\beta} \le \alpha$, because $\beta \le \omega^{\beta} \le \alpha$ by [L1]; it has a greatest element $\beta_0 = \bigcup B$, since $\bigcup B = 0$ forces $B = \{0\}$ and $0 \in B$, since $\bigcup B = \delta^{+}$ gives $\delta \in \beta$ for some $\beta \in B$ and hence $\delta^{+} \le \beta \le \bigcup B = \delta^{+}$ with $\beta \in B$, and since $\bigcup B = \lambda$ a limit gives $\omega^{\xi} < \omega^{\beta} \le \alpha$ for every $\xi \in \lambda$, whence $\omega^{\lambda} = \sup\{\omega^{\xi} : \xi \in \lambda\} \le \alpha$ and $\lambda \in B$; and then $\omega^{\beta_0} \le \alpha < \omega^{\beta_0^{+}}$, the second inequality because $\beta_0^{+} \notin B$.

Closure below a power of $\omega$: if $\mu < \omega^{\beta}$ and $\nu < \omega^{\beta}$ then $\mu + \nu < \mu + \omega^{\beta} = \omega^{\beta}$, by claim (b) of [L3] and step 2.1.

Existence, by induction on $\alpha > 0$: take $\beta_0$ from step 2.2, so $\omega^{\beta_0} \le \alpha < \omega^{\beta_0^{+}} = \omega^{\beta_0} \cdot \omega$; divide by $\omega^{\beta_0} > 0$ using [L2] to get $\alpha = \omega^{\beta_0} \cdot c_0 + \rho$ with $\rho < \omega^{\beta_0}$; here $c_0 \ne 0$, since $c_0 = 0$ would give $\alpha = \rho < \omega^{\beta_0} \le \alpha$, and $c_0 < \omega$, since $\omega \le c_0$ would give $\omega^{\beta_0} \cdot \omega \le \omega^{\beta_0} c_0 \le \alpha$ by claims (d) and (b) of [L3], contradicting $\alpha < \omega^{\beta_0} \cdot \omega$. If $\rho = 0$ then $\alpha = \omega^{\beta_0} c_0$ is a normal form of length $1$. Otherwise $0 < \rho < \omega^{\beta_0} \le \alpha$, so the claim at $\rho$ gives a normal-form datum for $\rho$ with leading exponent $\gamma_0$ and leading coefficient $d_0 \ge 1$, and $\omega^{\gamma_0} \le \omega^{\gamma_0} d_0 \le \rho < \omega^{\beta_0}$ by [L3], so $\gamma_0 < \beta_0$ by [L1] and [L6]; prefixing $(\beta_0, c_0)$ to that datum therefore yields a normal-form datum whose value is $\alpha$.

Tail bound: if $(\beta_i, c_i)_{i \in k}$ is a normal-form datum then the value $\tau$ of its tail $(\beta_i, c_i)_{1 \le i < k}$ satisfies $\tau < \omega^{\beta_0}$; indeed $\tau = 0 < \omega^{\beta_0}$ when $k = 1$, and for $i \ge 1$ each term satisfies $\omega^{\beta_i} c_i < \omega^{\beta_i} \cdot \omega = \omega^{\beta_i^{+}} \le \omega^{\beta_0}$ by claim (d) of [L3], [L1] and $\beta_i^{+} \le \beta_0$, so induction on the number of terms using step 3.1 gives $\tau < \omega^{\beta_0}$.

Uniqueness, by induction on $\alpha > 0$: let $(\beta_i, c_i)_{i \in k}$ be a normal-form datum of value $\alpha$, with tail value $\tau$, so that $\alpha = \omega^{\beta_0} c_0 + \tau$ with $\tau < \omega^{\beta_0}$ by step 4.1; then $\omega^{\beta_0} = \omega^{\beta_0} \cdot 1 \le \omega^{\beta_0} c_0 \le \alpha$ by [L3], and $\alpha < \omega^{\beta_0} c_0 + \omega^{\beta_0} = \omega^{\beta_0}(c_0 + 1) \le \omega^{\beta_0} \cdot \omega = \omega^{\beta_0^{+}}$ by [L3], [L4] and $c_0 + 1 \le \omega$; so $\omega^{\beta_0} \le \alpha < \omega^{\beta_0^{+}}$, which pins $\beta_0$ down, since a second datum with leading exponent $\gamma_0 \ne \beta_0$ would satisfy the same two inequalities and, say, $\beta_0 < \gamma_0$ would give $\alpha < \omega^{\beta_0^{+}} \le \omega^{\gamma_0} \le \alpha$ by [L1] and [L6]; with $\beta_0$ fixed, the two representations $\alpha = \omega^{\beta_0} c_0 + \tau = \omega^{\beta_0} d_0 + \sigma$ with $\tau, \sigma < \omega^{\beta_0}$ agree by the uniqueness in [L2], so $c_0 = d_0$ and $\tau = \sigma$; and $\tau < \omega^{\beta_0} \le \alpha$, so the claim at $\tau$ makes the two tails identical when $\tau > 0$, while $\tau = 0$ forces both data to have length $1$, since a tail of length at least $1$ has value at least $\omega^{\beta_1} c_1 > 0$.

Existence is step 3.2 and uniqueness is step 5.1, so every ordinal $\alpha > 0$ has exactly one Cantor normal form.