The more light, the less resistance there is through the photocell.
So, lets say you have a battery of 9v
You run 5v(depending on the threshold you choose for the amount of light) to the Positive(+) pole of the comparator. Then you would put the photocell in series
between the battery and the Negative(-) pole of the comparator. During the day or lighted conditions, the photocell will have less resistance therefore letting more voltage through to the comparator. During lighted conditions, that of course depending on the threshold you choose, the photocell will let more than 5v through to the negative poll, therefore the output bit would be Zero(0). During darker conditions, the photocell will resist the voltage more and thus less than 5v will get to the comparators negative poll resulting in the output bit being One(1) which, if im not mistaken, should be 5volts output. I guess that would depend on the source voltage though. Then just run that 5v directly to the LED but through a resistor so that only about 1volt gets to the LED to extend its life time... or 2 volts if you want it to be much brighter but not last as long.
Fixed the schematic