Proving that the hermitian conjugate of the product of two operators is the product of the two hermitian congugate operators in opposite order I have reach a step in a problem of my quantum mechanics textbook that requires me to prove the following.
$$\hat{A}=(\hat{Q}\hat{R})^{\dagger} = \hat{R}^{\dagger}\hat{Q}^{\dagger}$$
I tried to prove this by substituting $\hat{A}$ into $\langle\Psi_1|\hat{A} \Psi_2\rangle$
$$\Rightarrow \int_{-\infty}^\infty \mathrm{\Psi_1}(\hat{Q}\hat{R})^{\dagger}\Psi_2\,\mathrm{d}x$$
Do I integrate by parts next?
Any advice would be much appreciated
 A: You don't actually need to pick a basis as indicated in Andrew McAdams's answer.
This is easiest to prove in mathy notation (as opposed to Dirac notation) where $(\cdot, \cdot)$ is the inner product, then for all vectors $\phi$ and $\psi$ in the Hilbert space, and for operators $A$ and $B$, we have
\begin{align}
  (\phi, AB\psi) = (A^\dagger\phi, B\psi) = (B^\dagger A^\dagger\phi, \psi)
\end{align}
while on the other hand
\begin{align}
  (\phi, AB\psi) = ((AB)^\dagger\phi, \psi)
\end{align}
which implies $B^\dagger A^\dagger = (AB)^\dagger$ as desired. 
A: As leftaroundabout wrote, integration by parts is unuseful. You don't have the expressions for operators, so there is no reasoning for it. But you may use following:
\begin{align}
 \langle \Psi_{1}|(\hat {A}\hat {B})^{+} |\Psi_{2}\rangle 
& = \langle \Psi_{2} | \hat {A}\hat {B} |\Psi_{1}\rangle^{*} 
\\ & = \sum_{c}\langle \Psi_{2}| \hat {A}| c \rangle^{*} \langle c|\hat {B}| \Psi_{1}\rangle^{*} 
\\ & = \sum_{c}\langle c| \hat {A}^{+}| \Psi_{2} \rangle \langle \Psi_{1}|\hat {B}^{+}| c\rangle 
\\ & =\sum_{c}\langle \Psi_{1}| \hat {B}^{+}| c \rangle \langle c|\hat {A}^{+}| \Psi_{2}\rangle 
\\ & = \langle \Psi_{1}|\hat {B}^{+}\hat {A}^{+} |\Psi_{2}\rangle ,
\end{align}
where I used definition of hermitian conjugate,
$$
\langle \Psi_{1}| \hat {A}^{+}|\Psi_{2}\rangle = \langle \Psi_{2}| \hat {A}|\Psi_{1}\rangle^{*},
$$
and basis $|c\rangle $ of eigenvectors of an operator in a Hilbert space, $\langle c| c\rangle = 1$; $\sum_c|c\rangle\langle c|=\mathbb 1$
