It follows from the fact that the eignevalues of the hermitian operator \hat{n} = \hat{a}^{\dagger} \, \hat{a} are non-negative! Namely, let:
<br />
\hat{n} \, \vert \nu \rangle = \nu \, \vert \nu \rangle<br />
Assuming the eigenket \vert \nu \rangle is normalized to unity (\langle \nu \vert \nu \rangle = 1), we have:
<br />
\nu = \langle \nu \vert \hat{n} \vert \nu \rangle = \langle \nu \vert \hat{a}^{\dagger} \, \hat{a} \vert \nu \rangle<br />
Now, insert a compete orthonormal basis \lbrace k' \rbrace between \hat{a} and \hat{a}^\dagger:
<br />
\nu = \sum_{k'}{\langle \nu \vert \hat{a}^{\dagger} \vert k' \rangle \, \langle k' \vert \hat{a} \vert \nu \rangle} = \sum_{k'}{\left\vert \langle k' \vert \hat{a} \vert \nu \rangle \right\vert^2}<br />
Each of the summands is a non-zero real number, being a modulus of a complex number. Therefore, \nu \ge 0.
Now, from the commutation relation \left[ \hat{a}, \hat{a}^\dagger \right] = 1, it follows that the ket:
<br />
\hat{a} \, \vert \nu \rangle<br />
is also an eigenket of \hat{n}, corresponding to an eigenvalue \nu - 1. By inductive reasoning, we conclude that \nu - n, \forall n \in \mathbb{N} are also eigenvalues of the number operator.
So, let us suppose that N + 1 > \nu \ge N \ge 0, \ N \in \mathbb{N}. This means that \nu - N - 1 < 0, which is not allowed (because all the eigenvalues are non-negative!). The only way out is if \left( \hat{a} \right)^{N + 1} \, \vert \nu \rangle = 0 (not the ground state \vert 0 \rangle, but the zero ket in the Hilbert space!). But, then apply \left( \hat{a}^\dagger \right)^{N + 1} from the left, and use:
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<br />
\hat{X}_m \equiv \left[ \hat{a}^\dagger, \hat{a}^m \right]<br />
<br />
\hat{X}_{m + 1} = \left[ \hat{a}^\dagger, \hat{a} \, \hat{a}^m \right] = \hat{a} \, \left[ \hat{a}^\dagger, \hat{a}^m \right] + \left[ \hat{a}^\dagger, \hat{a} \right] \, \hat{a}^m = \hat{a} \, \hat{X}_m - \hat{a}^m<br />
<br />
\hat{X}_0 = 0<br />
<br />
\hat{X}_1 = \hat{a} \, 0 - \hat{1} = -\hat{1}<br />
<br />
\hat{X}_2 = \hat{a} \, (-\hat{1}) - \hat{a} = -2 \, \hat{a}<br />
...
Lemma:
<br />
\boxed{\hat{X}_m = -m \, \hat{a}^{m - 1}, \ m \ge 1}<br />
Proof:
<br />
\hat{X}_{m + 1} = \hat{a} \, \left(-m \, \hat{a}^{m - 1} \right) - \hat{a}^{m} = -(m + 1) \, \hat{a}^{m}<br />
By Principle of Mathematical Induction, Q.E.D.
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Then, we have:
<br />
\begin{array}{lcl}<br />
\left(\hat{a}^\dagger\right)^{N + 1} \, \hat{a}^{N + 1} \, \vert \nu \rangle & = & 0 \\<br />
<br />
\left( \hat{a}^\dagger \right)^N \, \hat{a}^\dagger \, \hat{a}^{N + 1} \, \vert \nu \rangle & = & 0 \\<br />
<br />
\left( \hat{a}^\dagger \right)^N \, \left( \hat{a}^{N + 1} \, \hat{a}^\dagger + X_{N + 1} \right) \, \vert \nu \rangle & = & 0 \\<br />
<br />
\left[ \left( \hat{a}^\dagger \right)^N \, \hat{a}^{N} \, \hat{a} \, \hat{a}^\dagger - (N + 1) \, \left( \hat{a}^\dagger\right )^{N} \, \hat{a}^{N} \right] \, \vert \nu \rangle & = & 0 \\<br />
<br />
\left[ \left( \hat{a}^\dagger \right)^N \, \hat{a}^{N} \left( \hat{n} + 1 \right) - (N + 1) \, \left( \hat{a}^\dagger \right)^N \, \hat{a}^{N} \right] \, \vert \nu \rangle & = & 0 \\<br />
<br />
\left( \nu - N \right) \, \left( \hat{a}^\dagger \right)^N \, \hat{a}^{N} \, \vert \nu \rangle & = & 0 \\<br />
<br />
\ldots & & \\<br />
<br />
\left( \nu - N \right) \, \left( \nu - N + 1\right) \, \ldots \nu \, \vert \nu \rangle & = & 0<br />
\end{array}<br />
Now, from the initial assumption \vert \nu \rangle \neq 0 (not a trivial eigenket), and N \le \nu < N + 1, we conclude that \nu = N.
This means that the only eigenvalues of the number operator \hat{n} = \hat{a}^\dagger \, \hat{a} are the non-negative integers. Q.E.D.
