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 $<\Psi_1|\hat{A} \Psi_2>$

$$\Rightarrow \int_{-\infty}^\infty \mathrm{\Psi_1}(\hat{Q}\hat{R})^{\dagger}\Psi_2\,\mathrm{d}x$$

Do I integrate by parts next?

Any advice would be much appreciated

-
No real need to do that, though you may be expected to. It's actually a far more basic identity than anything that would require an integral. You only need to shuffle the operators from side to side of the bra-ket expression, using the definition of the Hermitian conjugate. – leftaroundabout Oct 31 '13 at 18:00

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: $$\langle \Psi_{1}|(\hat {A}\hat {B})^{+} |\Psi_{2}\rangle = \langle \Psi_{2} | \hat {A}\hat {B} |\Psi_{1}\rangle^{*} = \sum_{c}\langle \Psi_{2}| \hat {A}| c \rangle^{*} \langle c|\hat {B}| \Psi_{1}\rangle^{*} = \sum_{c}\langle c| \hat {A}^{+}| \Psi_{2} \rangle \langle \Psi_{1}|\hat {B}^{+}| c\rangle =$$ $$=\sum_{c}\langle \Psi_{1}| \hat {B}^{+}| c \rangle \langle c|\hat {A}^{+}| \Psi_{2}\rangle^{*} = \langle \Psi_{1}|\hat {B}^{+}\hat {A}^{+} |\Psi_{2}\rangle ,$$ 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, $\langle c| c\rangle = 1$.

-

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.

-
and here as a single line, just for the heck of it: $((AB)^\dagger\phi, \psi) = (\phi, AB\psi) = (A^\dagger\phi, B\psi) = (B^\dagger A^\dagger\phi, \psi) \;\forall \phi,\psi \Leftrightarrow (AB)^\dagger = B^\dagger A^\dagger$ – Christoph Oct 31 '13 at 20:48