${}^\zeta$ Defining the vacuum state $\lvert 0 \rangle$ as that where $\hat a(\vec p)\lvert 0 \rangle=0$ and normalizing it as $\langle 0\lvert 0 \rangle = 1$, then we again build the Hilbert space as a Fock space from a basis of multiparticle states; $$\lvert \vec p_1,\vec p_2, \cdots,\vec p_n\rangle=\lvert\hat a(\vec p_1)^\dagger \hat a(\vec p_2)^\dagger\cdots\hat a(\vec p_n)^\dagger\lvert 0\rangle\quad \forall \vec p\tag{1}$$
The multiparticle states are orthogonal to each other if they differ in the particle momenta, and as a consequence the number of particles; suppose $\lvert \vec p_1,\vec p_2, \cdots,\vec p_n\rangle$ and $\lvert\vec q_1,\vec q_2, \cdots,\vec q_m\rangle$ are such that $\vec p_1\ne \vec q_i$ {ie. the first state has a particle with momentum $\vec p_1$ different to all those of the second}. Then, $$\langle \vec p_1,\vec p_2, \cdots,\vec p_n\lvert\vec q_1,\vec q_2, \cdots,\vec q_m\rangle$$
$$=\langle 0\lvert \hat a(\vec p_1)\hat a(\vec p_2)\cdots \hat a(\vec p_n)\hat a(\vec q_1)^\dagger \hat a(\vec q_2)^\dagger\cdots\hat a(\vec q_m)^\dagger\rvert 0 \rangle\tag{2}$$
$$=\langle 0\lvert \hat a(\vec p_2)\cdots \hat a(\vec p_n)\hat a(\vec q_1)^\dagger \hat a(\vec q_2)^\dagger\cdots\hat a(\vec q_m)^\dagger\hat a(\vec p_1)\rvert 0 \rangle=0$$
since $\hat a(\vec p_1)$ commutes with all the $\hat a^\dagger(\vec q_i)$ as they have different momenta, and $\hat a(\vec p_1)\rvert 0\rangle=0$.
I would like to understand why the first equality holds in equation $(2)$.
I know that the Hermitian adjoint, $\dagger$, acts as an invertible map between the bras and kets as elements of a Hilbert space, which, in Dirac notation would look like for states, $|\phi\rangle$, $|\psi\rangle$ $$|\phi\rangle=\hat O|\psi\rangle\overbrace{\iff}^\dagger\langle\phi|=\langle\psi|\hat O^\dagger$$
But the formulae above doesn't explain what happens if $\hat O$ is equal to a product of operators, say, $\hat O=\hat A \hat B \hat C$.
So starting with equation $(1)$, I take the Hermitian adjoint of both sides:
$$\left(\lvert \vec p_1,\vec p_2, \cdots,\vec p_n\rangle\right)^\dagger=\left(\lvert\hat a(\vec p_1)^\dagger \hat a(\vec p_2)^\dagger\cdots\hat a(\vec p_n)^\dagger\lvert 0\rangle\right)^\dagger$$
$$=\langle 0\lvert\left(\hat a(\vec p_1)^\dagger \hat a(\vec p_2)^\dagger\cdots\hat a(\vec p_n)^\dagger\right)^\dagger\rvert$$
$$=\langle 0\lvert\left(\hat a(\vec p_n)^\dagger\right)^\dagger\cdots\left(\hat a(\vec p_2)^\dagger\right)^\dagger \left(\hat a(\vec p_1)\right)^\dagger\rvert$$
$$\implies \langle \vec p_1,\vec p_2, \cdots,\vec p_n\rvert=\langle 0\lvert\hat a(\vec p_n)\cdots\hat a(\vec p_2)\hat a(\vec p_1)\rvert\tag{3}$$
Where in eqn. $(3)$ I assume that $$\hat O^\dagger=\left(\hat A^\dagger \hat B^\dagger \hat C^\dagger\right)^\dagger\stackrel{\color{red}{?}}{=} \left(\hat C^\dagger\right)^\dagger \left(\hat B^\dagger\right)^\dagger \left(\hat A^\dagger\right)^\dagger=\hat C \hat B \hat A\tag{4}$$
I put a question mark over the second equality in $(4)$ as after searching the internet I see all the time that $\left(\hat O_1 \hat O_2\right)^\dagger = \left(\hat O_2\right)^\dagger\left(\hat O_1\right)^\dagger$ but no evidence that it generalises to products of more that just two operators.
By analogy with $(1)$, I can write down a similar expression for $\lvert\vec q_1,\vec q_2, \cdots,\vec q_m\rangle$, $$\lvert\vec q_1,\vec q_2, \cdots,\vec q_m\rangle=\lvert\hat a(\vec q_1)^\dagger \hat a(\vec q_2)^\dagger\cdots\hat a(\vec q_m)^\dagger\lvert 0\rangle\quad \forall \vec q\tag{5}$$
If $(4)$ is true then multiplying together $(3)$ and $(5)$ to get the inner product, $\langle \vec p_1,\vec p_2, \cdots,\vec p_n\lvert\vec q_1,\vec q_2, \cdots,\vec q_m\rangle$ yields, $$\langle \vec p_1,\vec p_2, \cdots,\vec p_n\lvert\vec q_1,\vec q_2, \cdots,\vec q_m\rangle$$ $$=\langle 0\lvert\hat a(\vec p_n)\cdots\hat a(\vec p_2)\color{blue}{\hat a(\vec p_1)}\hat a(\vec q_1)^\dagger \hat a(\vec q_2)^\dagger\cdots\hat a(\vec q_m)^\dagger\lvert 0\rangle\ne (2)\tag{6}$$
In the last line of the quote it is stated that "since $\hat a(\vec p_1)$ commutes with all the $\hat a^\dagger(\vec q_i)$ as they have different momenta". But this statement only makes sense to me if the inner product is written as eqn. $(6)$ since the $\color{blue}{\hat a(\vec p_1)}$ commutes with all the creation operators, $\hat a(\vec q_i)^\dagger$ to the right of $\color{blue}{\hat a(\vec p_1)}$.
Since we are only told that "$\hat a(\vec p_1)$ commutes with all the $\hat a^\dagger(\vec q_i)$", what is the justification for commuting the $\hat a(\vec p_1)$ past all the other annihilation operators, $\hat a(\vec p_j)$, in the final equality of eqn. $(2)$?
In short, why did the the author not reverse the order of the operators in eqn. $(2)$ so that it is eqn. $(6)$?
${}^\zeta$ Notes from a QFT course at ICL dept. of physics.