= Solution
For either <number operator> $\hat N=\hat a^\dagger\hat a$, positivity follows from
$$
\langle u|\hat N|u\rangle=\|\hat a|u\rangle\|^2\ge0.
$$
Hence an <eigenvalue> $\nu$ cannot be negative. The <commutators>
$$
[\hat N,\hat a]=-\hat a,\qquad [\hat N,\hat a^\dagger]=\hat a^\dagger
$$
show that $\hat a|\nu\rangle$, if nonzero, is an <eigenstate> with <eigenvalue> $\nu-1$. For a normalized <eigenstate>, $\|\hat a|\nu\rangle\|^2=\nu$.
If $\nu$ were not an integer, successive nonzero lowerings would eventually reach an <eigenstate> with $0<\nu'<1$. Its next lowering would have nonzero <norm> but a negative <eigenvalue>, contradicting positivity. The ladder must therefore terminate at $\nu=0$, with $\hat a|0\rangle=0$. Conversely, successive applications of the <creation operator> never vanish, because
$$
\|\hat a^\dagger|n\rangle\|^2=\langle n|(\hat N+1)|n\rangle=n+1>0.
$$
The normalizable bottom state exists explicitly: solving $\hat a\psi_0=0$ gives $\psi_0(x)\propto\exp[-m\omega x^2/(2\hbar)]$. Thus the <number operator> has exactly the non-negative integer <eigenvalues>.
Since the two coordinate <number operators> commute, use the product <eigenstates> $|n_x,n_y\rangle$. Their <energy eigenvalues> are
$$
\boxed{E_{n_x,n_y}=\hbar\omega(n_x+n_y+1),\qquad n_x,n_y=0,1,2,\ldots.}
$$
At fixed $N=n_x+n_y$, the allowed pairs are $(0,N),(1,N-1),\ldots,(N,0)$. Thus the <energy level> $E_N=(N+1)\hbar\omega$ has <quantum degeneracy> $N+1$, and the <ground-state energy> is $\hbar\omega$.
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