I'm reading from a lecture note on introductory quantum mechanics (here), which says on P.3 that "The expectation value of any time-independent operator $\hat{Q}$ on a stationary state is time-independent". It proceeds to prove the statement in the following way: \begin{equation} \begin{split} \langle\hat{Q}\rangle_{\Psi}& \:=\int{dx\:\Psi^*(x,t)\hat{Q}\Psi(x,t)} \\ & = \int{dx\:e^{iEt/\hbar}\psi^*(x)\hat{Q}\psi(x)e^{-iEt/\hbar}}\\ & =\int{dx\:e^{iEt/\hbar}e^{-iEt/\hbar}\psi^*(x)\hat{Q}\psi(x)}\\ & = \langle\hat{Q}\rangle_{\psi} \end{split} \end{equation}
My Question is: why does the time-independence (i.e. no explicit time dependence) of the operator $\hat{Q}$ warrant the moving of the factor $e^{-iEt/\hbar}$ across the operator $\hat{Q}$? Why won't $\hat{Q}$ act on the factor even though itself is time-independent?