Indecomposable limit ordinals A limit ordinal $\gamma>0$ is said to be indecomposable iff $\nexists\alpha,\beta<\gamma$ such that $\alpha+\beta=\gamma$.  In view of this definition, I’m trying to prove the equivalence of the following statements for limit ordinals $\gamma$:
(1) $\gamma$ is indecomposable.
(2) $\alpha+\gamma=\gamma$ $\forall\alpha<\gamma$.
(3) $\gamma=\omega^\alpha$ for some $\alpha\in\mathrm{ON}$.
I’ve been able to get $(1)\implies(2)$ and $(3)\implies(1)$, but I have nothing more than scratch work for $(2)\implies(3)$.  Any suggestions would be appreciated.
 A: It's easy to show $(1)\implies(3)$ and $(2)\implies(1)$, and deduce the final implication. Because both of these are really obvious.
To see $(1)\implies(3)$ note that if $\gamma$ is indecomposable then its Cantor normal form has to have only one term, and it has to be $\omega^\alpha$ for some $\alpha$.
To see that $(2)\implies(1)$ note that if $\alpha,\beta<\gamma$ then we have that $\alpha+\beta+\gamma=\gamma$, but if $\alpha+\beta=\gamma$ then $\gamma+\gamma=\gamma$, which is impossible.
A: Suppose that $\gamma$ is not a power of $\omega$. Let $A=\{\alpha\le\gamma:\alpha=\omega^\beta\text{ for some }\beta\in\mathbf{ON}\}$, and let $B=\{\beta\in\mathbf{ON}:\omega^\beta\in A\}$; then $\sup A=\omega^{\sup B}$, so $\sup A<\gamma$. Let $\alpha=\sup A$ and $\beta=\sup B$, and let $\eta$ be the unique ordinal such that $\alpha+\eta=\gamma$. Suppose that $\eta=\gamma$; then $\alpha+\gamma=\gamma$, and by an easy induction $\alpha\cdot n+\gamma=\gamma$ for each $n\in\omega$. But then $\alpha\cdot n\le\gamma$ for each $n\in\omega$, and therefore 
$$\omega^{\beta+1}=\omega^\beta\cdot\omega=\alpha\cdot\omega=\sup_{n\in\omega}\alpha\cdot n\le\gamma<\omega^{\beta+1}\;,$$
an obvious contradiction.
A: (*1) {A limit ordinal $\gamma>0$ is indecomposable}.
(*2) {$\gamma>0, \alpha +\gamma=\gamma, \forall \alpha<\gamma$}.
(*3) {$\gamma>0, \exists \alpha, \gamma=\omega^\alpha$}.
Now we prove {(*1) $\Rightarrow$  (*2) $\Rightarrow$  (*3) $\Rightarrow$  (*1).}
{{(*1) $\Rightarrow$  (*2):} $\forall \alpha,\beta<\gamma$, we have $\alpha+\beta<\gamma$ (}{otherwise if $\exists \alpha,\beta<\gamma, \alpha+\beta>\gamma$ we should have $\exists \beta'<\beta, \alpha+\beta'=\gamma$ . Therefore, we know  given arbitrary $\alpha<\gamma$, then $\forall \beta<\gamma, \alpha+\beta<\gamma$ which implies $\gamma\geq \lim\limits_{\beta\rightarrow \gamma} \alpha+\beta=\alpha+\gamma$, and we know $\alpha+\gamma\geq \gamma$ and thus we know $\gamma=\alpha+\gamma, \forall \alpha<\gamma$.
{{(*2) $\Rightarrow$  (*3):} By Cantor's normal form theorem and virtrue of its proof we know $\gamma=\sum_{i=1}^n \omega^{\beta_i}\cdot k_i=\omega^{\beta_1}\cdot k_1+\rho$ where $\rho<\gamma$ and $k_1<\omega$ and thus by (*2) we  should have $\gamma=\omega^{\beta_1}\cdot k_1$. Now suppose $k_1\geq 2$ then we know $\exists k'<\omega, k_1=k'+1$ and thus $\gamma=\omega^{\beta_1}\cdot k_1=\omega^{\beta_1}\cdot k'+\omega^{\beta_1}$ which contradicts (*2) once notice that $\omega^{\beta_1}\cdot k', \omega^{\beta_1}< \gamma=\omega^{\beta_1}\cdot k_1$. Therefore, we know $k_1\leq 1$ and given $\gamma>0$ we know $k_1=1$ and thus $\gamma=\omega^{\beta_1}$ which implies $\exists \alpha, \gamma=\omega^{\alpha}$.
{{(*3) $\Rightarrow$  (*1):} Given $\exists \alpha, \gamma=\omega^{\alpha}$ and $\gamma>0$, suppose $\exists m,n<\omega^\alpha, \omega^\alpha=m+n$. Apply Cantor's normal form theorem to $m,n$, we would have $m=\sum_{i=0}^{l_m}\omega^{\beta^m_i}\cdot k^m_i, n=\sum_{j=0}^{l_n}\omega^{\beta^n_i}\cdot k^n_i$, then we can rearrange these two summations and have $\omega^\alpha=\sum_{i=0}^{l_m}\omega^{\beta^m_i}\cdot k^m_i + \sum_{j=0}^{l_n}\omega^{\beta^n_i}\cdot k^n_i =\sum_{l=0}^{L}\omega^{\beta_l}\cdot k_l$ where $\left\{ \beta_l:l<L \right\}=\left\{ \beta^n_i: i< l_n+1 \right\} \cup \left\{ \beta^m_i: i< l_m+1 \right\}$ and length $L\leq l_m+l_n+2$ is finite and $\beta_0=\max\left\{ \left\{ \beta^n_i: i< l_n+1 \right\} \cup \left\{ \beta^m_i: i< l_m+1 \right\}  \right\}$,  and for the rest $\beta_l$ we have $\beta_l=\max\left\{ \left\{ \beta^n_i: i< l_n+1 \right\} \cup \left\{ \beta^m_i: i< l_m+1 \right\}\setminus \left\{\beta_{l'},  l'<l \right\}  \right\}, \forall L>l\geq 1$ and $k_l=\sum_{i=0}^{l_m}k_i^m\cdot \mathbf{1}_{\{\beta^m_i=\beta_l\}}+\sum_{j=0}^{l_n}k_j^n\cdot \mathbf{1}_{\{\beta^n_j=\beta_l\}}, \forall l<L$. Notice $\omega^{\beta_0}\geq \max\{m,n\}<\omega^\alpha$ and thus this representation contradicts the uniqueness stated in Cantor's normal form theorem and thus we know $\forall m,n<\omega^\alpha=\gamma, \alpha+\beta\neq \gamma$, which implies limit ordinal $\gamma>0$ is indecomposable.
