How do we prove that capacitance depends only on the geometric properties of the conductors? I was wondering if the statement that the capacitance of a capacitor depends only the the geometric properties of the conductors forming the capacitor has a mathematically rigorous proof(when there is no dielectric medium placed in between)?
 A: While The Photon is correct in saying that it also depends on the material between them, I think that it is common practice to first show in classes that a capacitor's C in a vacuum is dependent only on it's geometry. Then to move on to looking at dielectrics. 
The reason they do this is because using that fact makes it FAR easier to do calculations about capacitance. It simplifies a lot of things, especially because in introductory courses you usually use very symmetrical capacitors. Therefore the geometry is easy to work with. It's more useful to focus on the physics of what's happening rather than wasting class time doing the math every time.
As to whether or not there is a mathematical proof... probably. If you can't find one, then you can always attempt one yourself.
A: 
the capacitance of a capacitor depends only the the geometric properties of the conductors forming the capacitor

This isn't true. 
It also depends on the dielectric properties of the material between the conductors.
Since it's not true, it can't be proven, mathematically or otherwise.
A: As other answers have pointed out this result would be for the vacuum case. I can sketch the proof for the "simplest scenario" and then you can look for the general case. This is when $N$ conductors are enclosed inside a big $N+1$ conductor surface. The $N+1$ surface is used so we can easily resort to unicity of solutions. We will also assume the $N$ small conductors have no cavities for simplicity, although this case can be extrapolated once we finish.
Let's denote with $S_i$ the surface of each conductor, and $S_{N+1}$ will be the inner enclosing surface of the big conductor. We will also denote with $\vec{n}_i$ the normal pointing outwards of each conductor, while $\vec{n}_{N+1}$ points inwards the surface of the big enclosing conductor, towards the small conductors. Let's denote with $S_T$ the composed surface of all surfaces $S_1,S_2$,... , $S_{N+1}$. Let's call $V_T$ the volume enclosed by $S_T$.
Each $S_i$ will be an equipotential surface, i.e. $\Phi(S_i) = \phi_i$. The potential in $V_T$ must satisfy the Laplace equation with the former as boundary conditions.
The trick is now to use linearity of the solutions. For $\Phi$ in $V_T$, we should be able to write
\begin{equation}
\Phi = \sum^{N+1}_{j=1} \phi_j f_j \, ,
\end{equation}
where the $f_i$ are functions that satisfy the Laplace equation in $V_T$ and satisfy the boundary conditions $f_i (S_j) = \delta_{ij}$. Unicity says each $f_i$ and $\Phi$ are unique.
Now it just remains to notice that the total charge in each conductor will be given by
\begin{equation}
Q_i = \int_{S_i} \sigma_i dS = - \epsilon_0 \int_{S_i} \nabla \Phi \cdot \vec{n}_i dS = - \epsilon_0 \sum^{N+1}_{j=1} \phi_j  \int_{S_i} \nabla f_j \cdot \vec{n}_i dS = \sum^{N+1}_{j=1} \phi_j C_{ij} \, ,
\end{equation}
from which one finds that the
\begin{equation}
C_{ij} = - \epsilon_0 \int_{S_i} \nabla f_j \cdot \vec{n}_i dS
\end{equation}
depend only on the geometry.
A: When you check the mathematical definition of the capacitance of a finite conductor, you will find that this statement is correct in vacuum.
