Why do wavefunctions for stationary states include $e^{-iEt/\hbar}$? Stationary states are separable solutions with $\Psi(x, t)=\psi(x)e^{-iEt/\hbar}$. But why is that there? Griffiths (Section 2.1 Stationary states, equation 2.8) says that observables for these states are constant in time, so the time-dependent factor can be left out. So why do we include that factor?
Unlike the duplicate, I want to ask "Can we safely assume $\Psi(x,t)=\psi(x)$ and simply abandon that whole $\exp$ altogether?"
 A: The time dependent Schroedinger equation looks like this:
$$ i\hbar \frac{\partial \Psi}{\partial t} = H \Psi = \left ( -\frac{\hbar^2 }{2 m}\frac{\partial^2}{\partial x^2} + V(x,t) \right ) \Psi(x,t) ,$$
you attempt a solution via separation of variables: $\Psi(x,t) = \psi(x) T(t)$, plug it in.
If the potential $V$ is time independent such that $V(x,t) = V(x)$, then the equation above splits into two independent equations:
$$\left ( -\frac{\hbar^2 }{2 m}\frac{\mathrm{d}^2}{\mathrm{d} x^2} + V(x) \right ) \psi(x) = E \psi(x), \quad \text{Time independent Schroedinger equation} $$
and:
$$ i\hbar\frac{\mathrm{d} T}{\mathrm{d} t} = ET \implies T(t) \propto e^{-i\frac{Et}{\hbar}} = e^{-i\omega t}.$$
With $E$ a constant identified with energy.
Hence the full solution will be $\Psi(x,t) = \psi(x) e^{-i\omega t}$.
A: It does not affect observables if the system is in an eigenstate. However, as soon as you have a superposition of states with different energies, the amplitudes will oscillate depending on the energy difference because of this factor.
A: The time independent Schrödinger equation is derived from the time dependent Schrödinger equation by applying a separation of variables: $\Psi(\vec{r},t)=\psi(\vec{r})f(t)$. 
Applying this result into the time dependent Schrödinger equation,
$$i\hbar\frac{f'(t)}{f(t)}=\left(-\frac{\hbar^2}{2m}\nabla^2\psi+V(\vec{r})\psi\right)/\psi$$
Since the left part of the equation is a function of $t$, and the right part of the equation is a function of $\vec{r}$, we can conclude that either side of the equation must be equal to a constant, which we will call $E$.
The equation corresponding to $f(t)$ is $i\hbar f'+Ef=$, so $f(t)=e^{-iEt/\hbar}$.
The other equation is the time independent Schrödinger equation. Once you calculate the eigenstates and eigenvalues of the time independent Schrödinger equation, you can write the time evolution of the eigenstates:
$$\Psi(\vec{r},t)=\psi(\vec{r})e^{-iEt/\hbar}$$
A: Noah is right. Actually, his (or her) answer is related with the concept of phase factor. When you have a wave function:
$$\psi(x,t)$$
if you add a global phase factor, for example:
$$\widetilde{\psi}(x,t)=e^{i\theta}\psi(x,t)$$
the physical consecuences of both wave functions: $\psi$ and $\widetilde{\psi}$ are the same. On the other hand, if your wave function is a superposition like this:
$$\psi(x,t)=\lambda_1\psi_1(x,t)+\lambda_2\psi_2(x,t)$$
with $\psi_n(x,t)=e^{iE_n t/\hbar}\phi_n(x)$
the phase factor of every single function $\psi_n(x,t)$ contributes to the total wave, producing an interference term.
