Hypsometric equation derivation: why is g/R in the exponent?

[arenaninja]

Homework Statement

Derive the hypsometric equation, assuming a sea level temperature of 15 C, and that the temperature decreases with heigh at a rate of 6.5 C per km.

Homework Equations

Ideal gas law:
$$P=\rho RT$$
Hydrostatic equilibrium:
$$dP = -\rho gdz$$
Temperature varies with height:
$$T = 288 - 6.5*10^{-3}z$$

The Attempt at a Solution

$$\int_{P_{0}}^{P} \frac{dP}{P} = \int_{0}^{z} \frac {-gdz}{RT(z)}$$
$$\int_{P_{0}}^{P} \frac{dP}{P} = \int_{0}^{z} \frac {-gdz}{R*(288 - 6.5*10^{-3}z)}$$
$$ln\left(\frac{P}{P_{0}}\right)=\frac{g}{6.5*10^{-3}*R}ln\left(\frac{288-6.5*10^{-3}z}{288}\right)$$
And exponentiating this:
$$\frac{P}{P_{0}} = e^{\frac{g}{0.0065R}}\left(1-2.26x10^{-5}z\right)$$
and finally
$$P = P_{0}\left(1-2.26x10^{-5}z\right)e^{\frac{g}{0.0065R}}$$

but from a dimensional analysis, I know g/R is not a dimensionless quantity and therefore I should *not* be exponentiating this. I expect an exponential in the final answer but I'm not sure whether this is correct. (Some of you may recognize the initial conditions set this to be the equation used by a standard altimeter).

Did I make any conceptual mistakes? I'm fairly confident about the math.

 

[MathematicalPhysicist]

The constant 6.5*10^-3 has units as well.