I didn't see this either but...
It looks like you have a 100K resistor in series with your LED. At 2ma current draw that implies 240 volts is present.
That's a Mains Neon not an LED. They have a strike voltage usually about 90V and for UK 240 Volt AC Mains it's common to see them with a 220K series resistor, so i'm guessing here that 100K is probably for 110V AC Mains.
As it looks like it's a Mains voltage device one option could be to use an opto-isolator. They are available as AC input versions (back to back led's inside) but because they only have LED's inside they operate at about 2V so you will need a suitable series resistor. 100K will probably do the job BUT the wattage is crucial when dropping Mains down to only a couple of volts to run an LED. For instance 5mA to drive an LED via 100K would in theory need a resistor rated at 2.5 Watts or more and it would increase to 10 watts @ 10mA. They do get very hot when dropping that much voltage even at low currents.
An opto-isolator as it's name implies, would give you isolation from the high voltages but i don't like the idea of all that heat generated in the series resistor.
There are other ways to go about this, it's possible to drop the voltage through an X2 rated capacitor or use a small transformer or back to basics with a good old fashioned Mains AC relay and use the contacts to drive the input of the dev board.