 Solution

The map $\beta \mapsto {\omega }^{\beta }$ is strictly increasing and continuous at limit ordinals. Also ${\omega }^{\beta }\ge \beta$, so the set $S=\{\beta :{\omega }^{\beta }\le \alpha \}$ is nonempty and bounded by $\alpha$. If $\sup S$ is a successor, it is already in $S$; if it is a limit, continuity gives ${\omega }^{\sup S}={\sup }_{\beta \in S}{\omega }^{\beta }\le \alpha$. Thus $S$ has a greatest element $\beta$.

The initial segment of the well-order $\alpha$ of order type ${\omega }^{\beta }$ leaves a tail of some order type $\gamma$, giving ${\omega }^{\beta }+\gamma =\alpha$. For fixed left summand, ordinal addition is strictly increasing in its right argument, so this $\gamma$ is unique.

Maximality gives $\alpha <{\omega }^{\beta +1}={\omega }^{\beta }\omega$. Choose the least positive integer $m$ such that $\alpha <{\omega }^{\beta }m$. Then $m\ge 2$ and ${\omega }^{\beta }(m-1)\le \alpha$. If $\gamma \ge {\omega }^{\beta }(m-1)$, monotonicity of ordinal addition would give $\alpha ={\omega }^{\beta }+\gamma \ge {\omega }^{\beta }m$, a contradiction. Hence $\gamma <{\omega }^{\beta }(m-1)\le \alpha$. If $\alpha$ is not a power of $\omega$, both ${\omega }^{\beta }$ and $\gamma$ are nonzero and smaller than $\alpha$, proving decomposability.

Conversely, prove by transfinite induction that any two ordinals below ${\omega }^{\beta }$ have sum below ${\omega }^{\beta }$. For $\beta =0$ the only smaller ordinal is zero. For $\beta =\delta +1$, choose finite $m,n$ with $x<{\omega }^{\delta }m$ and $y<{\omega }^{\delta }n$. Monotonicity gives $x+y<{\omega }^{\delta }(m+n)<{\omega }^{\delta +1}$. For limit $\beta$, choose $\delta <\beta$ sufficiently large that $x,y<{\omega }^{\delta }$, and apply the induction hypothesis. Thus the nonzero indecomposable ordinals are exactly the powers of $\omega$.