Memory is shaky, I think this covers it.
Current is constant because no charge accumulates anywhere in the circuit. Over time, as much charge leaves a region as enters it. If there's no entering and leaving, there's no current.
At it's most fundamental, Ohm's Law asserts that $\vec{J}=\sigma\vec{E}$, i.e. The current density vector is proportional to the applied electric field, where that constant of proportionality is the conductance. If we assume that electric field is constant, as implied by constant current and Ohm's Law, then it must be the negative of the potential difference, divided by the distance between field points. In other words, the potential difference causes the constant electric field and not the resistance.
Now $\vec{J}$ is on average in the direction of the current with magnitude $I/A$ where $I$ is current, and $A$ is cross-sectional area of the circuit. The electric field also points in the direction of the current and has magnitude $V/l$ where $V$ is the potential difference between two points in the circuit and $l$ is the length between them.
So:
$\vec{J}=\sigma \vec{E}\implies \frac{I}{A}=\sigma\frac{V}{l}\implies I=\frac{\sigma A}{l} V$.
The reciprocal of the conductance($\sigma$) is the resistivity($\rho$) and we have $R=\frac{\rho l}{A}$ and $V=IR$.
Given $\vec{J}=\sigma \vec{E}$, take the divergence of both sides.
By the continuity equation for charge, we have that $\nabla \cdot \vec{J}=-\partial\rho/\partial t$ where $\rho$ is charge density. By Gauss' Law, $\nabla \cdot \vec{E}=\rho/\epsilon_0$.
So: $-\partial \rho/\partial t = \frac{\sigma}{\epsilon_0}\rho$
So $\rho=\rho_0e^{-\sigma t/\epsilon_0}$ by solving the differential equation.
So within the wire net charge exponentially decays when there's an applied field.
The foregoing analysis fails on the surface and in non-Ohmic materials, but for some reason that isn't covered before grad school.
