Clarification of multipole expansion for a point charge In Griffith's electrodynamic: 3.4.2 He pointed out that the monopole term is the exact potential for a single point charge. 
However I was under the impression that different configuration of a charge distribution can act as a point charge from superposition thus allowing other multipoles to exist? 
If not how do I prove that a single point charge only has monopole?
 A: The multipole coefficients associated with a $1/|r|$ distribution $\rho$ depends on the choice of origin. For example, if you have a point charge and you choose the origin to be at that point charge, then it will have a pure monopole character. However, if you choose the origin to be elsewhere, it will have nonzero expansion coefficients other than the monopole. This is an artifact of your choice of coordinate system.
To make this rigorous, let $\mathbf{I}=\{I_0^0,I_1^{-1},I_1^{0},I_1^1,...\}$ and $\mathbf{R}=\{R_0^0,R_1^{-1},R_1^{0},R_1^1,...\}$ be the set of irregular and regular solid harmonics. Then the potential $V(\mathbf{r})$ due to $\rho$ admits the exterior and interior multipole expansions
$$V=\sum_{J=0}^\infty \mathbf{I}_J\rangle\left\langle\mathbf{R}_J,\rho\right\rangle\qquad\text{when }|\mathbf{r}|>r_\text{max}
\\
V=\sum_{J=0}^\infty \mathbf{R}_J\rangle\left\langle\mathbf{I}_J,\rho\right\rangle\qquad\text{when }|\mathbf{r}|<r_\text{min}
$$
or in matrix notation,
$$V=\mathbf{I}\mathbf{R}^\dagger\rho\qquad\text{when }|\mathbf{r}|>r_\text{max}
\\
V=\mathbf{R}\mathbf{I}^\dagger\rho\qquad\text{when }|\mathbf{r}|<r_\text{min}.$$
In the case where $\rho$ is purely real, we can use the real solid harmonics $\mathbf{I}'$ and $\mathbf{R}'$, which are related to the standard solid harmonics by a unitary block diagonal matrix $\mathbf{U}$ via $\mathbf{I}'=\mathbf{I}\mathbf{U}$ from which we obtain the analogous real expansions
$$V=\mathbf{I}\mathbf{R}^\dagger\rho=\mathbf{I}\mathbf{U}\mathbf{U}^\dagger\mathbf{R}^\dagger\rho=[\mathbf{I}'][\mathbf{R}']^\dagger\rho=[\mathbf{I}'][\mathbf{R}']^\mathsf{T}\rho\qquad\text{when }|\mathbf{r}|>r_\text{max}
\\
V=\mathbf{R}\mathbf{I}^\dagger\rho=\mathbf{R}\mathbf{U}\mathbf{U}^\dagger\mathbf{I}^\dagger\rho=[\mathbf{R}'][\mathbf{I}']^\dagger\rho=[\mathbf{R}'][\mathbf{I}']^\mathsf{T}\rho\qquad\text{when }|\mathbf{r}|<r_\text{min}$$
which has the advantage that the list of multipole moments $[\mathbf{I}']^\mathsf{T}\rho$ or $[\mathbf{R}']^\mathsf{T}\rho$ are purely real.
So, why does a point charge not located at the origin have moments other than a monopole? It's for the same reason why a washing machine with a raccoon inside of it will shake around when it's on a wash cycle: it's not balanced, as the charges (or mass) are not located at the center of the relevant coordinate system. 
As an explicit proof of why a point charge not located at the origin can't have a pure monopole moment, suppose otherwise. Then a test charge will be uniformly accelerated towards the center of the coordinate system, instead of towards the point charge. This is a contradiction. Therefore, there must be higher moments involved.
Alternatively,  a detailed justification can also be obtained by applying the addition theorem for spherical harmonics, but hopefully the proof given in the previous paragraph is sufficiently illuminating to show why higher moments will appear when a point charge is not located at the chosen origin.
Here's a numerical example to compute the moments of a single point charge located at spherical coordinate $(R,\pi/2,0)$ in Mathematica (it also computes the potential $V$ at an arbitrary point and compares it to the potential obtained from direct application of $V=1/|\mathbf{r}-\mathbf{r}_0|$):
SolidHarmonicI[l_, m_, r_, \[Theta]_, \[Phi]_] := 
  Sqrt[(4 \[Pi])/(2 l + 1)]
    SphericalHarmonicY[l, m, \[Theta], \[Phi]]/r^(l + 1);
SolidHarmonicR[l_, m_, r_, \[Theta]_, \[Phi]_] := 
  Sqrt[(4 \[Pi])/(2 l + 1)] r^
   l SphericalHarmonicY[l, m, \[Theta], \[Phi]];
SphToCart = CoordinateTransform["Spherical" -> "Cartesian", #] &;
r = {R, \[Pi]/2, 0};(*Spherical coordinates of point charge*)

Q[L_, m_] := ((-1)^m SolidHarmonicR[L, -m, ##] & @@ 
    r) q;(*Exterior multipole moment or order (L,m)*)
MatrixForm[
 Table[Q[L, m], {L, 0, 5}, {m, -L, L}]]
rule = {R -> 1.2, q -> 2, 
   rtest -> 5.2, \[Theta]test -> 1.2, \[Phi]test -> 2.3};
Chop[Sum[SolidHarmonicI[L, m, rtest, \[Theta]test, \[Phi]test] Q[L, 
     m], {L, 0, 5}, {m, -L, L}] /. rule]
q/Norm[SphToCart@r - 
    SphToCart@{rtest, \[Theta]test, \[Phi]test}] /. rule


0.332219
0.332273
$$\left(
\begin{array}{c}
 \{q\} \\
 \left\{\frac{q R}{\sqrt{2}},0,-\frac{q R}{\sqrt{2}}\right\} \\
 \left\{\frac{1}{2} \sqrt{\frac{3}{2}} q R^2,0,-\frac{q R^2}{2},0,\frac{1}{2} \sqrt{\frac{3}{2}} q R^2\right\}
   \\
 \left\{\frac{1}{4} \sqrt{5} q R^3,0,-\frac{1}{4} \sqrt{3} q R^3,0,\frac{1}{4} \sqrt{3} q R^3,0,-\frac{1}{4}
   \sqrt{5} q R^3\right\} \\
 \left\{\frac{1}{8} \sqrt{\frac{35}{2}} q R^4,0,-\frac{1}{4} \sqrt{\frac{5}{2}} q R^4,0,\frac{3 q
   R^4}{8},0,-\frac{1}{4} \sqrt{\frac{5}{2}} q R^4,0,\frac{1}{8} \sqrt{\frac{35}{2}} q R^4\right\} \\
\end{array}
\right)$$

Note that there are nonzero moments of all orders whenever $R\neq 0$. However, the potential at the test location is correct up to parts per thousand accuracy when the sum runs up to $L=4$.

how do I prove that a single point charge only has monopole?

Set $R=0$ in the above triangle of numbers. Everything vanishes except the monopole term.
