Relationship between rate of reaction and pressure Given the equation

$$\ce{A(g) + B(g) → C(g) + D(g)}$$  

with reaction rate given by

$$v = k[\ce{A}]^2$$

How can I prove that reaction rate goes up with pressure? I tried substituting $k$ for $Ae^{-E_A/(RT)}$ which leaves me with

$$v = Ae^{-\frac{E_A}{RT}}[A]^2$$

If I use

$$PV = nRT \implies RT = \frac{1}{n}PV \implies RT = \frac{P}{C}$$

I end up with this: 

$$v = Ae^{-E_A\frac{C}{P}}[A]^2$$

This shows there is a relation, but I don't know what to do with the concentration $C$, or whether it was correct to plug in the ideal gas law in the first place.  Any help?
The exponential term in the Arrhenius equation does not imply reaction with other molecules. It applies to ready to react unstable adduct, about to overcome the energy barrier $E_a$ to form the product.
The concentration of this adduct is driven by concentration of reagents and therefore by the pressure.

Therefore the reaction rate equation implicitly involve 2 multiplicative terms:


Concentration of reagent adduct

$$\mathrm{const}\cdot [A]^2 = \mathrm{const}\cdot \left(\frac{p_{\rm A}}{RT}\right)^2$$
Probability the adduct reaches the reaction energy barrier, from the Boltzmann distribution:

$$\exp{\left(-\frac{E_A}{RT}\right)}$$


By other words, the constant $A$ in your equation

$$v = A \exp\left(-\frac{E_A}{RT}\right)$$

implicitly involves concentration, respectively pressure of the reagents.