= Solution
Let $\psi$ have energy $E_n$ and put $\gamma=2m\hbar\omega$. The lowering <norm> formula gives, inductively,
$$
\|(a_-)^j\psi\|^2=\gamma^j\frac{n!}{(n-j)!}\|\psi\|^2,\qquad 0\leq j\leq n.
$$
In particular $(a_-)^n\psi$ is a nonzero <ground state>. By the assumed nondegeneracy, it equals $c\psi_0$ for some $c\ne0$.
It remains to show that raising this vector recovers $\psi$, rather than merely another state at the same energy. On an <energy eigenstate> at $E_n$, factorization gives $a_+a_-\psi=\gamma n\psi$. Iterating the factorization down the ladder yields
$$
(a_+)^n(a_-)^n\psi=\gamma^n n!\,\psi.
$$
Applying $(a_+)^n$ to $(a_-)^n\psi=c\psi_0$ therefore proves
$$
\boxed{\psi=\frac{c}{(2m\hbar\omega)^n n!}(a_+)^n\psi_0.}
$$
For $n=0$ this is the original ground-state assumption. This proves <nondegeneracy propagated by an oscillator ladder>: every energy <eigenspace> is one-dimensional if the <ground state> is nondegenerate.
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