Why particle number operator $\hat{N}$ is $\hat{a}^\dagger\hat{a}$ rather than $\hat{a}\hat{a}^\dagger$? Both $\hat{a}^\dagger\hat{a}$ and $\hat{a}\hat{a}^\dagger$ are Hermitian, how do we know which one represents the particle number?
 A: Since you define e.g. in the bosonic case
$c_j^\dagger: H_N^S \rightarrow H_{N+1}^S,\quad   c_j^\dagger | \ldots n_j \ldots \rangle := \sqrt{n_j+1} |\ldots n_j+1 \ldots \rangle$
$c_j: H_N^S \rightarrow H_{N-1}^S,\quad  c_j | \ldots n_j \ldots \rangle := \sqrt{n_j} |\ldots n_j-1 \ldots \rangle$
it makes more sence to use $a^\dagger a$ which will give you $n_j$ (instead of $n_j+1$) as prefactor when acting on a state $| \ldots n_j \ldots \rangle $.
A: The vacuum should have particle number $0$. In some detail: we would like $\hat{N}\ |0\rangle=0$ and $\hat{N}=a^\dagger a$ is the only ordering that does that. It follows from the usual commutation relations that $\hat{N} |n\rangle =n\ |n\rangle$ which is in sync with interpretation of  $|n\rangle $ as an $n$-particle state in second-quantisation.
A: We require that the number operator have the following property:
$$\hat n |0\rangle = 0.$$
We know that 
$$\hat a |0\rangle = 0$$
and we know that
$$\hat a |1\rangle = |0\rangle$$
and we know that
$$\hat a^{\dagger} |0\rangle = |1\rangle. $$
Thus, it follows that
$$\hat a \hat a^{\dagger} \ne \hat n$$
since
$$\hat a \hat a^{\dagger} |0\rangle = \hat a|1\rangle \ne 0.$$
Now, it remains to be shown that
$$\hat a^{\dagger}\hat a  = \hat n. $$
Can you take it from here?
