Is there charge build up before a resistor? I understand that Kirchhoff's current law says that the current, $I$, is constant throughout a resistor, i.e. there is no build up of charge in a resistor. All charge going in to the resistor is the same as all charge coming out. In other words, Coulombs/sec going in = Coulombs/sec coming out. 
However, suppose we have an ideal wire, i.e. a wire with no resistance (or a physical wire with very little resistance; but lets use an ideal one), if there is a battery providing a voltage, the current through the wire is infinite. Once we get to the resistor the current is a finite amount $I = dq/dt = V/R,$ in other words, the current has decreased. So, from what I understand there should be a little charge pileup at the entrance to the resistor, and in fact, this is what gives the resistor voltage to drive the current through. Is this true? If not, please explain. 
 A: The voltage that drives the current is different in different parts of the circuit. The sum of the voltages driving current through each part of the circuit, wire and resistor must be equal to the voltage applied with a power source or battery.
In the case of the circuit with the resistor and wire then nearly all the voltage will be pushing electrons through the resistor and only a tiny bit will be pushing the electrons through the wire. If the resistance of the wire is zero then there would be no voltage pushing electrons through the wire - they would just flow on their own (as electrons flow in a superconductor).
Charge does sometimes pile up, but we call that capacitance, as I am sure you know. I mention it here because there can be 'stray' unintended capacitance (and also inductance) which people need to worry about when they use really high frequency AC circuits in the high RF or microwave region.
So it is possible that charge would build up in front of a resistor, but this would be due to stray capacitance mostly but in answer to your question....
Final point, as you know the voltage drop is over the resistor so the potential is different at each end of the resistor. It will be that the surface of the wire on different sides of the resistor will have a slightly different charge density and thus slightly different potential, thus there will be a slight difference in charge density each side of the resistor. I think that this pile up of charge would be very tiny though, but you are correct about this. 
To estimate this final effect one could make a guess at the capacitance of the wire and then use $Q=CV$ to determine the difference in charge between the two wires. I would make a guess that we are talking about C of less than $10^{-12}~F$ - if say $10^{-15}~F$ and $10 V$ the pile up would be about $10^{-14}~Coulombs$ or about $100,000$ electrons.
** Final edit **
the capacitance of a wire depends on its length, radius and distance to nearest 'earth' the value chosen above may be on the small side, but I expect the capacitance would be  less than $10^{-12}~F$. See here to calculate wire capacitance.
A: Crudely, electrons repel each other and even out the charge. While the influence of the electrons travels at a good fraction of the speed of light the electrons themselves do not move much. From this link
" The electricity that is conducted through copper wires in your home consists of moving electrons. The protons and neutrons of the copper atoms do not move. The actual progression of the individual electrons in a given direction through the wire is quite slow. The electrons have to work their way through the billions of atoms in the wire and this takes considerable time. In the case of a 12 gauge copper wire carrying 10 amperes of current (typical of home wiring), the individual electrons only move about 0.02 cm per sec or 1.2 inches per minute (in science this is called the drift velocity of the electrons.). "
