[proofplan] We prove existence by strong induction on $\alpha < \varepsilon_0$. Throughout, juxtaposition $\omega^\beta c$ means ordinal multiplication on the right by the positive integer $c$. For a nonzero ordinal $\alpha$, we choose the largest exponent $\beta$ with $\omega^\beta \leq \alpha$, then choose the largest positive integer $c$ with $\omega^\beta c \leq \alpha$; an ordinal interval lemma gives a unique remainder below $\omega^\beta$, so induction gives a normal form with strictly smaller exponents. For uniqueness, we show that the leading term $\omega^\beta c$ is determined by the interval containing $\alpha$, then use the same interval lemma to identify the common remainder and repeat on the smaller tail. [/proofplan] [step:Choose the leading exponent and coefficient] Let $\alpha$ be a nonzero ordinal with $\alpha < \varepsilon_0$. Define the set of admissible leading exponents \begin{align*} E_\alpha := \{\gamma < \varepsilon_0 : \omega^\gamma \leq \alpha\}. \end{align*} The set $E_\alpha$ is nonempty because $0 \in E_\alpha$, since $\omega^0 = 1 \leq \alpha$. It is bounded above by $\varepsilon_0$. Let \begin{align*} \beta := \sup E_\alpha. \end{align*} First we verify that $\beta < \varepsilon_0$. If $\beta = \varepsilon_0$, then continuity of ordinal exponentiation at the limit ordinal $\varepsilon_0$ and the defining fixed-point identity $\omega^{\varepsilon_0}=\varepsilon_0$ give \begin{align*} \varepsilon_0 = \omega^{\varepsilon_0} = \sup_{\gamma < \varepsilon_0} \omega^\gamma \leq \alpha, \end{align*} contradicting $\alpha < \varepsilon_0$. Thus $\beta < \varepsilon_0$. Ordinal exponentiation with base $\omega$ is increasing and continuous in the exponent. If $\beta$ is a limit ordinal, then \begin{align*} \omega^\beta = \sup_{\gamma < \beta} \omega^\gamma \leq \alpha, \end{align*} because every $\gamma < \beta = \sup E_\alpha$ is bounded by some element of $E_\alpha$ and hence $\omega^\gamma \leq \alpha$. If $\beta$ is a successor or $0$, the same conclusion follows from the definition of supremum together with monotonicity. Hence $\beta \in E_\alpha$, so $\beta$ is the largest ordinal satisfying $\omega^\beta \leq \alpha$. Now define \begin{align*} C_\alpha := \{m \in \mathbb{N} : \omega^\beta m \leq \alpha\}. \end{align*} This set is nonempty because $1 \in C_\alpha$. It is finite: if $\omega^\beta m \leq \alpha$ for every $m \in \mathbb{N}$, then by continuity of ordinal multiplication in the right argument, \begin{align*} \omega^{\beta+1} = \omega^\beta \omega = \sup_{m \in \mathbb{N}} \omega^\beta m \leq \alpha, \end{align*} contradicting maximality of $\beta$, since then $\beta+1 \in E_\alpha$. Therefore $C_\alpha$ has a largest element. Denote it by \begin{align*} c := \max C_\alpha. \end{align*} By definition, \begin{align*} \omega^\beta c \leq \alpha < \omega^\beta(c+1). \end{align*} We now use the following ordinal interval lemma. [claim:Represent ordinals in one additively principal interval] Let $\lambda$ be an ordinal and let $\kappa$ be a nonzero ordinal. For every ordinal $\eta$ satisfying \begin{align*} \lambda \leq \eta < \lambda + \kappa, \end{align*} there is a unique ordinal $\xi < \kappa$ such that \begin{align*} \eta = \lambda + \xi. \end{align*} [/claim] [proof] The map \begin{align*} f: \kappa &\to [\lambda,\lambda+\kappa) \\ \xi &\mapsto \lambda + \xi \end{align*} is the canonical order embedding of the second summand in the ordinal sum $\lambda+\kappa$: by the definition of ordinal addition as ordered concatenation, the final block of order type $\kappa$ is exactly the interval \begin{align*} [\lambda,\lambda+\kappa). \end{align*} Therefore every $\eta$ in this interval has the form $\lambda+\xi$ for a unique $\xi<\kappa$. Uniqueness also follows from strict monotonicity of ordinal addition in the right argument: if $\xi_0<\xi_1<\kappa$, then $\lambda+\xi_0<\lambda+\xi_1$. [/proof] Apply the claim with $\lambda := \omega^\beta c$, $\kappa := \omega^\beta$, and $\eta := \alpha$. Since ordinal multiplication by the positive integer $c+1$ on the right satisfies \begin{align*} \omega^\beta(c+1)=\omega^\beta c+\omega^\beta, \end{align*} the interval condition gives a unique ordinal $\rho<\omega^\beta$ such that \begin{align*} \alpha = \omega^\beta c + \rho. \end{align*} [guided] We first isolate the largest possible power of $\omega$ that can appear at the front of the normal form. For the given nonzero ordinal $\alpha < \varepsilon_0$, define \begin{align*} E_\alpha := \{\gamma < \varepsilon_0 : \omega^\gamma \leq \alpha\}. \end{align*} This set is nonempty because $\omega^0 = 1$, and every nonzero ordinal is at least $1$. Thus $0 \in E_\alpha$. We set \begin{align*} \beta := \sup E_\alpha. \end{align*} Before using $\beta$ as an exponent below $\varepsilon_0$, we verify that it is not equal to $\varepsilon_0$. If $\beta=\varepsilon_0$, then continuity of ordinal exponentiation at the limit ordinal $\varepsilon_0$, together with $\omega^{\varepsilon_0}=\varepsilon_0$, gives \begin{align*} \varepsilon_0 = \omega^{\varepsilon_0} = \sup_{\gamma < \varepsilon_0}\omega^\gamma \leq \alpha, \end{align*} which contradicts the hypothesis $\alpha<\varepsilon_0$. Thus $\beta<\varepsilon_0$. The point is that $\beta$ is not merely a supremum; it is actually attained. This uses the standard continuity of ordinal exponentiation in the exponent: \begin{align*} \omega^\lambda = \sup_{\gamma < \lambda} \omega^\gamma \end{align*} for every limit ordinal $\lambda$, together with monotonicity in the successor and zero cases. Since every exponent below $\beta=\sup E_\alpha$ is bounded by an admissible exponent, the corresponding power of $\omega$ is at most $\alpha$. Therefore $\omega^\beta \leq \alpha$, so $\beta \in E_\alpha$. Hence $\beta$ is the largest exponent with $\omega^\beta \leq \alpha$. Next we choose the largest finite coefficient of this leading power. Define \begin{align*} C_\alpha := \{m \in \mathbb{N} : \omega^\beta m \leq \alpha\}. \end{align*} Again $C_\alpha$ is nonempty because $1 \in C_\alpha$. It cannot contain every natural number. If it did, then \begin{align*} \omega^{\beta+1} = \omega^\beta \omega = \sup_{m \in \mathbb{N}} \omega^\beta m \leq \alpha. \end{align*} That would imply $\beta+1 \in E_\alpha$, contradicting the fact that $\beta$ is the largest element of $E_\alpha$. Hence $C_\alpha$ is a nonempty finite subset of $\mathbb{N}$, so it has a maximum. Call this maximum $c$. The maximality of $c$ gives the interval condition \begin{align*} \omega^\beta c \leq \alpha < \omega^\beta(c+1). \end{align*} We now need a genuine ordinal interval fact, not ordinary algebraic subtraction. For ordinals $\lambda$ and nonzero $\kappa$, the definition of the ordinal sum $\lambda+\kappa$ as an ordered concatenation says that the final block of order type $\kappa$ is exactly the interval \begin{align*} [\lambda,\lambda+\kappa). \end{align*} Consequently every $\eta$ in this interval has a unique representation $\eta=\lambda+\xi$ with $\xi<\kappa$; uniqueness follows because ordinal addition is strictly increasing in the right argument. Apply this with \begin{align*} \lambda := \omega^\beta c, \qquad \kappa := \omega^\beta, \qquad \eta := \alpha. \end{align*} Since \begin{align*} \omega^\beta(c+1)=\omega^\beta c+\omega^\beta, \end{align*} the interval condition produces a unique ordinal $\rho<\omega^\beta$ such that \begin{align*} \alpha = \omega^\beta c + \rho. \end{align*} [/guided] [/step] [step:Construct the Cantor normal form by strong induction] We prove existence by strong induction on nonzero ordinals $\alpha < \varepsilon_0$. Assume that every nonzero ordinal $\delta < \alpha$ has a Cantor normal form. Apply the preceding step to obtain ordinals $\beta,\rho$ and a positive integer $c$ such that \begin{align*} \alpha = \omega^\beta c + \rho, \qquad \rho < \omega^\beta. \end{align*} If $\rho = 0$, then \begin{align*} \alpha = \omega^\beta c \end{align*} is already a Cantor normal form. If $\rho > 0$, then $\rho < \alpha$, because $\alpha = \omega^\beta c + \rho$ and $\omega^\beta c > 0$. By the induction hypothesis, there exist $r \in \mathbb{N}$, ordinals \begin{align*} \gamma_1 > \gamma_2 > \cdots > \gamma_r \geq 0, \end{align*} and positive integers $d_1,\dots,d_r$ such that \begin{align*} \rho = \omega^{\gamma_1} d_1 + \omega^{\gamma_2} d_2 + \cdots + \omega^{\gamma_r} d_r. \end{align*} Since $\rho < \omega^\beta$, the leading exponent of $\rho$ satisfies $\gamma_1 < \beta$; otherwise $\omega^{\gamma_1} \leq \rho$ would imply $\omega^\beta \leq \rho$. Therefore \begin{align*} \alpha = \omega^\beta c + \omega^{\gamma_1} d_1 + \omega^{\gamma_2} d_2 + \cdots + \omega^{\gamma_r} d_r \end{align*} has strictly decreasing exponents. This is a Cantor normal form for $\alpha$. [guided] We now turn the leading-term construction into a full normal form. The proof is by strong induction: assume that every nonzero ordinal $\delta < \alpha$ already has a Cantor normal form, and prove the same for $\alpha$. From the previous step, we have a largest leading exponent $\beta$, a largest positive coefficient $c$, and a remainder $\rho$ satisfying \begin{align*} \alpha = \omega^\beta c + \rho, \qquad \rho < \omega^\beta. \end{align*} If $\rho = 0$, then there is nothing more to decompose: \begin{align*} \alpha = \omega^\beta c \end{align*} is a valid Cantor normal form with one term. Suppose instead that $\rho > 0$. Since $\omega^\beta c > 0$ and \begin{align*} \alpha = \omega^\beta c + \rho, \end{align*} the remainder $\rho$ is strictly smaller than $\alpha$. Hence the strong induction hypothesis applies to $\rho$. Thus there exist a natural number $r$, ordinals \begin{align*} \gamma_1 > \gamma_2 > \cdots > \gamma_r \geq 0, \end{align*} and positive integers $d_1,\dots,d_r$ such that \begin{align*} \rho = \omega^{\gamma_1} d_1 + \omega^{\gamma_2} d_2 + \cdots + \omega^{\gamma_r} d_r. \end{align*} We must check that placing $\omega^\beta c$ in front keeps the exponents strictly decreasing. Since $\rho < \omega^\beta$, the leading exponent $\gamma_1$ of $\rho$ must satisfy $\gamma_1 < \beta$. If $\gamma_1 \geq \beta$, then monotonicity of ordinal exponentiation would give \begin{align*} \omega^\beta \leq \omega^{\gamma_1} \leq \rho, \end{align*} contradicting $\rho < \omega^\beta$. Therefore $\beta > \gamma_1$, and the concatenated expression \begin{align*} \alpha = \omega^\beta c + \omega^{\gamma_1} d_1 + \omega^{\gamma_2} d_2 + \cdots + \omega^{\gamma_r} d_r \end{align*} has strictly decreasing exponents and positive integer coefficients. This is the desired Cantor normal form. [/guided] [/step] [step:Show the leading exponent and coefficient are forced] Suppose \begin{align*} \alpha = \omega^{\beta_1} c_1 + \omega^{\beta_2} c_2 + \cdots + \omega^{\beta_k} c_k \end{align*} is a Cantor normal form, where \begin{align*} \beta_1 > \beta_2 > \cdots > \beta_k \geq 0 \end{align*} and each $c_i$ is a positive integer. Let \begin{align*} \rho := \omega^{\beta_2} c_2 + \cdots + \omega^{\beta_k} c_k \end{align*} if $k \geq 2$, and let $\rho := 0$ if $k=1$. Since every exponent appearing in $\rho$ is smaller than $\beta_1$, we have \begin{align*} \rho < \omega^{\beta_1}. \end{align*} Therefore \begin{align*} \omega^{\beta_1} c_1 \leq \alpha < \omega^{\beta_1}(c_1+1). \end{align*} It follows that $\beta_1$ is the largest ordinal $\beta$ such that $\omega^\beta \leq \alpha$: if $\beta>\beta_1$, then the ordinal successor property gives $\beta_1+1\leq\beta$, and monotonicity of ordinal exponentiation gives \begin{align*} \omega^{\beta_1+1}\leq \omega^\beta. \end{align*} But \begin{align*} \alpha < \omega^{\beta_1}(c_1+1) < \omega^{\beta_1}\omega = \omega^{\beta_1+1}, \end{align*} so no such $\beta$ can satisfy $\omega^\beta\leq\alpha$. Once $\beta_1$ is fixed, the interval inequality shows that $c_1$ is the largest positive integer $c$ such that $\omega^{\beta_1} c \leq \alpha$. Hence the leading exponent and leading coefficient are uniquely determined by $\alpha$. [guided] Assume that $\alpha$ is written in Cantor normal form: \begin{align*} \alpha = \omega^{\beta_1} c_1 + \omega^{\beta_2} c_2 + \cdots + \omega^{\beta_k} c_k, \end{align*} with \begin{align*} \beta_1 > \beta_2 > \cdots > \beta_k \geq 0. \end{align*} We isolate the tail after the first term. Define \begin{align*} \rho := \omega^{\beta_2} c_2 + \cdots + \omega^{\beta_k} c_k \end{align*} when $k \geq 2$, and define $\rho := 0$ when $k=1$. The important observation is that the tail is smaller than the leading power. Since each exponent in the tail is strictly below $\beta_1$, every term of the tail is below $\omega^{\beta_1}$, and the finite sum of terms with leading exponent below $\beta_1$ is still below $\omega^{\beta_1}$. Hence \begin{align*} \rho < \omega^{\beta_1}. \end{align*} Thus \begin{align*} \alpha = \omega^{\beta_1} c_1 + \rho \end{align*} lies in the ordinal interval \begin{align*} \omega^{\beta_1} c_1 \leq \alpha < \omega^{\beta_1} c_1 + \omega^{\beta_1} = \omega^{\beta_1}(c_1+1). \end{align*} This interval condition forces the leading exponent and coefficient. If some larger exponent $\beta > \beta_1$ satisfied $\omega^\beta \leq \alpha$, then the successor property of ordinal order gives $\beta_1+1\leq\beta$: every ordinal strictly above $\beta_1$ is at least the immediate successor $\beta_1+1$. By monotonicity of ordinal exponentiation, \begin{align*} \omega^{\beta_1+1} \leq \omega^\beta \leq \alpha. \end{align*} This contradicts \begin{align*} \alpha < \omega^{\beta_1}(c_1+1) < \omega^{\beta_1}\omega = \omega^{\beta_1+1}. \end{align*} So $\beta_1$ is the largest possible leading exponent. Once $\beta_1$ is fixed, the inequalities \begin{align*} \omega^{\beta_1} c_1 \leq \alpha < \omega^{\beta_1}(c_1+1) \end{align*} say exactly that $c_1$ is the largest positive integer $c$ with $\omega^{\beta_1}c \leq \alpha$. [/guided] [/step] [step:Cancel the common leading term and finish uniqueness by induction] We prove uniqueness by strong induction on $\alpha$. Suppose $\alpha$ has two Cantor normal forms: \begin{align*} \alpha &= \omega^{\beta_1} c_1 + \omega^{\beta_2} c_2 + \cdots + \omega^{\beta_k} c_k, \\ \alpha &= \omega^{\gamma_1} d_1 + \omega^{\gamma_2} d_2 + \cdots + \omega^{\gamma_\ell} d_\ell. \end{align*} By the previous step, the leading exponent and leading coefficient are determined by $\alpha$. Hence \begin{align*} \beta_1 = \gamma_1, \qquad c_1 = d_1. \end{align*} Define the two remainders \begin{align*} \rho &:= \omega^{\beta_2} c_2 + \cdots + \omega^{\beta_k} c_k, \\ \sigma &:= \omega^{\gamma_2} d_2 + \cdots + \omega^{\gamma_\ell} d_\ell, \end{align*} with the convention that an empty sum is $0$. Since \begin{align*} \omega^{\beta_1} c_1 + \rho = \omega^{\beta_1} c_1 + \sigma \end{align*} and both $\rho,\sigma < \omega^{\beta_1}$, the interval representation claim from the leading-term step applies with $\lambda:=\omega^{\beta_1}c_1$ and $\kappa:=\omega^{\beta_1}$. Its uniqueness part gives $\rho=\sigma$. If $\rho=\sigma=0$, then $k=\ell=1$, and the two normal forms agree. If $\rho=\sigma>0$, then $\rho<\alpha$, so the induction hypothesis applied to $\rho$ shows that the two tails agree term by term: \begin{align*} k=\ell, \qquad \beta_i=\gamma_i, \qquad c_i=d_i \end{align*} for every $2 \leq i \leq k$. Together with the equality of the leading terms, the two Cantor normal forms of $\alpha$ are identical. Therefore every nonzero ordinal $\alpha<\varepsilon_0$ has exactly one Cantor normal form. [guided] We now prove uniqueness. The method is again strong induction on $\alpha$: assume that every nonzero ordinal below $\alpha$ has at most one Cantor normal form, and suppose that $\alpha$ has two such expressions: \begin{align*} \alpha &= \omega^{\beta_1} c_1 + \omega^{\beta_2} c_2 + \cdots + \omega^{\beta_k} c_k, \\ \alpha &= \omega^{\gamma_1} d_1 + \omega^{\gamma_2} d_2 + \cdots + \omega^{\gamma_\ell} d_\ell. \end{align*} The previous step says that the leading exponent and coefficient are not choices; they are determined by the interval in which $\alpha$ lies. Therefore \begin{align*} \beta_1 = \gamma_1, \qquad c_1 = d_1. \end{align*} Now remove the common leading term. Define \begin{align*} \rho &:= \omega^{\beta_2} c_2 + \cdots + \omega^{\beta_k} c_k, \\ \sigma &:= \omega^{\gamma_2} d_2 + \cdots + \omega^{\gamma_\ell} d_\ell, \end{align*} where an empty sum means $0$. The equality of the two normal forms becomes \begin{align*} \omega^{\beta_1} c_1 + \rho = \omega^{\beta_1} c_1 + \sigma. \end{align*} Both remainders satisfy \begin{align*} \rho < \omega^{\beta_1}, \qquad \sigma < \omega^{\beta_1}, \end{align*} because all their exponents are strictly smaller than $\beta_1$. This is exactly the situation covered by the interval representation claim proved in the leading-term step: with $\lambda:=\omega^{\beta_1}c_1$ and $\kappa:=\omega^{\beta_1}$, an ordinal in \begin{align*} [\omega^{\beta_1}c_1,\omega^{\beta_1}(c_1+1)) = [\lambda,\lambda+\kappa) \end{align*} has a unique expression $\lambda+\xi$ with $\xi<\kappa$. Applying uniqueness to the common ordinal \begin{align*} \omega^{\beta_1} c_1 + \rho = \omega^{\beta_1} c_1 + \sigma \end{align*} gives $\rho=\sigma$. If this common remainder is $0$, then both normal forms consist only of the leading term, so $k=\ell=1$ and the forms agree. If the common remainder is nonzero, then it is strictly smaller than $\alpha$, and the induction hypothesis applies to it. The two tails are Cantor normal forms of the same ordinal $\rho$, so they agree term by term: \begin{align*} k=\ell, \qquad \beta_i=\gamma_i, \qquad c_i=d_i \end{align*} for every $2 \leq i \leq k$. Combining this tail equality with the already established equality of the leading exponent and coefficient proves that the two original Cantor normal forms are identical. Thus the Cantor normal form of every nonzero ordinal below $\varepsilon_0$ is unique. [/guided] [/step]