quantum mechanics - Why must the eigenvalue of the number operator be an integer?

My apologies if this question has already been answered. Using the notation of Sakurai, we have an energy eigenket N and an eigenvalue n.

$$N | n \rangle = n | n \rangle$$

For the number operator $N = a^\dagger a $. My question is precisely why does n have to be an integer. I do understand that it must be positive definite due to the norm condition on $ a | n \rangle $:

$$\langle n | a^\dagger a | n \rangle \geq 0$$

I can see that we can show:

$$ a | n \rangle = \sqrt{n-1} | n-1 \rangle $$ and similarly for $a^\dagger | n \rangle$

Sakurai argues that the sequential operation of $a$ operators leads to the form which he does not write, but I will write here for clarity:

$$ a^k | n \rangle = \left(\sqrt{n}\sqrt{n-1}...\sqrt{n-k+1}\right) | n - k + 1 \rangle $$

Since

$$ a|0 \rangle = 0 $$

the sequence must terminate, but can only terminate if n is an integer. How can we make that jump? Is there a formal mathematical proof that the sequence $\left(\sqrt{n}\sqrt{n-1}...\sqrt{n-k+1}\right)$ will terminate if and only if n is an integer?

Answer

1. 
   

   Show that if $|\nu\rangle$ is an eigenvector of $N$ with eigenvalue $\nu$, and if $a|\nu\rangle\neq 0$, then $a|\nu\rangle$ is an eigenvector of $N$ with eigenvalue $\nu-1$.

   
   


2. 
   

   Convince yourself that the spectrum of the number operator is non-negative.

   
   

Assume, by way of contradiction, that there is some non-integer eigenvalue $\nu^*>0$ and let $m$ denote the smallest integer larger than $\nu^*$. Use property 1 repeatedly ($m$-times) to show that $a^m|\nu^*\rangle$ is an eigenvector of $N$ with eigenvalue $\nu^*-m <0$. This is a contradiction QED.