Can I remove the LED from a pro mini and still use the raw pin to power it?
that would be unadvisable, but it could be done and the arduino would still work.
Didnt you have another thread on this matter? Keep related questions in same thread. But yes, you can. Be careful if trying to cut traces, because the trace may be routed suck that you cut or short something. Better to use a hot iron. I usually use pliers to get off the plastic over the die, and then rest iron across it with some added solder. But could also probably just rest iron across top and let it melt through plastic. Either works
No, do not bridge the pins. That will make it consume more power than it did with the LED (it will consume about 5/3rds as much as the LED did, approximately)
As long as you only remove the power LED, that will not have any effect on whether you can power via the RAW pin. That said, if you're powering through the raw pin, you're not all that concerned with power efficiency from the get-go...
Removing the regulator, of course, will make the RAW pin inoperable (the regulator is the thing that generates the 5v from the input on RAW). Usually the assumption is that when when you're trying to reduce power consumption, you're running with batteries within the appropriate voltage range tied directly to Vcc, ie, two (at 8mhz) or three (if you want 16mhz) AA's or a single cell LiPo (8mhz is in spec, 16mhz is not, but typically works anyway, at least at room temperature. Might not at the high end of the temperature range) - I suspect if you had read the article that suggested removing the regulator, it would have also described supplying power direct to vcc.
rayluc:
Do you think that I could use 6 volts on the vcc pin after removing the regulator on the 16mtz
No, absolute max Vcc is 6v, and fully charged alkaline battery is a little more than 1.5v per cell, and recomended operating condition is max 5.5v.
What about running it through a silicon diode and a schottky diode (in series, between the +6v and Vcc pin, band towards arduino board)? That would give a drop of approx 1v (0.7v for the silicon diode, 0.3v for the schottky)...
Actually, thinking about this now, I might be tempted to use 2 silicon diodes, for 1.4v drop, to keep you safer with fully charged batteries, or those Lithium-AA replacement batteries. Alkaline starts off at 1.65 under no load when brand new, though even a few % of capacity expended, or a light load, they're around 1.5 or slightly below that. Lithium AA replacement batteries start off with 1.8v/cell (!!) though rapidly fall off to 1.5~1.6. So your fresh 4S alkaline battery pack would start off at 6.6v (5.2v after the diodes), and fresh lithiums would be 7.2v (5.8 after the diodes, still higher than you'd like), with 6v (4.6v after diodes) or 6.4v (5.0 after diodes) over most of their discharge curve - those numbers assume no/virtually no load.
You pay for it on how drained the batteries can be before the arduino quits working, but in practice, a 5v/16MHz pro mini will generally keep limping along until voltage gets down to mid 3.x volts, often as low as 3v, at least at room temp (though external hardware it's communicating with might not like that, depending on what it is - what else is involved here?) - by which point the batteries would be pretty much dead anyway. This would become a problem faster of course if you're using it with something that puts a load on the batteries, as the voltage drops a little under load, or a lot under a very heavy, intermittent load that would drop the battery voltage while the heavy load was running, thus potentially resetting the arduino... some experimentation with the system at hand is probably in order. At least diodes are extremely abundant, and cheap as dirt (1.7cents each shipped from china for 1N4001 on ebay in lots of 100, or in kits with variety of normal and schottky diodes for $5-12, depending on how many of each diode, and whether it ships domestic or from china) 
As I may have mentioned before, I would bring this in through the 6-pin FTDI header, so that you can't possibly connect a serial adapter to it with external power also connected - you don't want the voltage on the vcc pin of the serial adapter plugged into your computer to be higher than the voltage on your computer's 5v rail - this will usually make the computer panic and shut down, or shut off the port, but that requires having faith that the computer's USB protection circuit really works well, which I wouldn't want to rely on. With the diodes, it's safe if the voltage is lower than computer's 5v rail, but in this case it would not be the case if the batteries were fresh.
You could also rig up an external LDL1117 5.0v regulator (awkward to solder, since it's an SMD part - but the pins are large enough that you could use "dead-bug" wiring style without too much pain) - that has a quiescent current of ~0.25mA, and minimum dropout of 0.3v (not a typo!), while IIRC the one used on a nano is like 5mA... though if you're sleeping the arduino this would probably still be a dealbreaker, since the power consumption of the arduino itself in deepsleep is 1uA (0.001mA) if you do everything right.