Why does current lead voltage in a capacitive circuit?

Imagine that you start with no charge in the capacitor in question and then connect it to some kind of source. For the purpose of argument, lets suggest that it is a battery and the current is flowing through a resistor.

At the instant the battery is connected, the capacitor has no charge on it and thus has zero potential across it (zero volts). So there is the full battery voltage across the resistance and maximum current flows.

After a small period of time, some current will have flowed and the capacitor will have reached a potential somewhere between zero and the battery voltage. At this point, there will be a smaller potential across the resistance and so less current will flow. The current will be falling and the voltage across the capacitor will be rising.

At some further point in time, the sufficient current will have flowed to increase the charge on the capacitor such that the potential across it is almost equal to that across the battery. The voltage will now be at maximum. Because of this, there will be almost no potential across the resistance and so almost no current will flow, current will be at minimum.

That is sufficient for a DC circuit, but with AC there is more to it. Consider the above to approximate to what happens on the first quarter-cycle of AC. From then on, the supply voltage will fall and current will gradually start to flow through the resistor, in reverse from above, as the capacitor voltage is now greater than the supply voltage. At some point in time, the supply voltage will have reached the zero crossing-point and the capacitor will be discharging with maximum current flow.

It should now be apparent that one can extrapolate the above for a complete cycle of AC and thence for a continuous AC supply.