= Solution
The map $\beta\mapsto\omega^\beta$ is strictly increasing and continuous at limit <ordinals>. Also $\omega^\beta\geq\beta$, so the set $S=\{\beta:\omega^\beta\leq\alpha\}$ is nonempty and bounded by $\alpha$. If $\sup S$ is a successor, it is already in $S$; if it is a limit, continuity gives $\omega^{\sup S}=\sup_{\beta\in S}\omega^\beta\leq\alpha$. Thus \b[$S$ 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 $m\geq2$ and $\omega^\beta(m-1)\leq\alpha$. If $\gamma\geq\omega^\beta(m-1)$, monotonicity of <ordinal addition> would give $\alpha=\omega^\beta+\gamma\geq\omega^\beta m$, a contradiction. Hence \b[$\gamma<\omega^\beta(m-1)\leq\alpha$.] 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 \b[the nonzero indecomposable ordinals are exactly the powers of $\omega$.]
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