Do we need to expand the potential in a power series to show $[x, V(x)] = 0$? Today in class (Intro to QM) we went over a couple of commutators. Among them was $[x, V]$, where $V=V(x)$ is a potential. What the teacher said to prove this is zero was: let's assume $V$ is analytic and can be expanded in a power series. Then we can take the commutator of $x$ and each term in the series, and since $[x, x^n] = 0$ everything is zero.
I immediately thought of something much simpler: In coordinate space, both $x$ and $V$ are simply multiplication. For any function $\psi$, $xV\psi = Vx\psi$ trivially, because we're just taking products. Therefore, they commute.
Is there anything wrong with my reasoning?
 A: There is nothing wrong with your reasoning; the issue is that your starting assumption that the potential can be treated as a multiplication operator needs to be justified.  Here's why:
Let's consider the potential operator $V$ acting on a state $|\psi\rangle$.  Let $X$ denote the position operator.  The definition of the position space representation of $V$ acting on the position space wavefunction $\psi$ is as follows:
\begin{align}
  V\psi(x) = \langle x| V|\psi\rangle
\end{align}
As you note, if we could somehow show that there exists some function $\tilde V$ such that
\begin{align}
  V\psi(x) = \tilde V(x)\psi(x) \tag{$\star$}
\end{align}
then we would have
\begin{align}
  XV\psi(x) &= X(\tilde V(x)\psi)(x) = \tilde V(x)X\psi(x) = \tilde V(x) x\psi(x)
\end{align}
while we would also have
\begin{align}
  VX\psi(x) = V(x\psi)(x) = xV\psi(x) = x\tilde V(x)\psi(x) = \tilde V(x)x\psi(x)
\end{align}
which would give $XV = VX$ as desired.  The issue is that that property $(\star)$ does not come for free from the definition above it.  If, however, $V$ is, for example, defined to be some analytic function $\tilde V$ of the position operator, then that property (modulo some mathy subtleties) does hold because for any positive integer power $n$ of $X$, one has
\begin{align}
  X^n\psi(x)=\langle x|X^n|\psi\rangle = x^n\langle x|\psi\rangle = x^n\psi(x)
\end{align}
by acting $X$ on the left repeatedly.  One then uses this for each term in the power series expansion of $V$ in $X$ to obtain the property $(\star)$.
Note. I am abusing notation here slightly and using the same symbol $V$ for the position operator and its position space representation.
A: I agree with joshphysic's answer that your reasoning is right, but you need to justify $V(x)$'s being a multiplication operator (in position space, that is).
Maybe the way I tend to look at this may help: I find in matters like these physical motivations tend to be the most intellectually fulfilling.
You are presumably dealing with a first quantized wave equation with a "semiclassical" model of interaction with the outside world  - say the Schrödinger or Dirac equation with a conservative potential impressed on the quantum particle's Hamiltonian by a central, approximately unshifting charged nucleus. The quantum nature of the origins of this potential are ignored for simplicity - otherwise we would be dealing with a quantum many body problem. 
So once the eigenvalue $x$ of the position observable $X$ has been measured by applying the observable to the quantum particle's state, the standard quantum postulates say that the particle must be in the position eigenstate corresponding to $x$. So at the instant just after the measurement, the particle's position is certain, and so the only sensible value we can postulate for the "potential energy" measurement is $v(x)$, where $v(x)$ is the classical potential function, and the value of the classical potential is certain for the instant after the measurement, once we know the position is $x$.
Likewise for any other position measurement with position eigenstate.
Therefore we see that every position eigenstate is an eigenstate of the observable we need to build for the semiclassical potential, and since the position eigenstates are complete, i.e. any quantum state is a superposition of these states, then we see that we have just fully defined the diagonalisation of the potential energy observable. Namely, it is diagonal in position co-ordinates, i.e. it is the multiplication operator $V \psi(x) = v(x) \psi(x)$.
Your main result also follows from considering the fact that position eigenstates make the semiclassical potential certain, i.e. they are all eigenstates and the only eigenstates for the potential. Given suitable reasonable assumptions about the operators and quantum state space concerned, operators commute if and only if they have the same eigenstates.
