"Efficiency" of Switching Power Supplies and Theoretical Energy Loss
This is a much simplified answer: -

There is this thing called a transformer - it has two coils that are closely magnetically connected. The two coils are called primary and secondary. If the primary has 1000 turns and the secondary has 100 turns it will step-down an AC voltage of 120V to 12V. There will be a slight power loss but usually transformers are better than 90% efficient. If we assumed near 100 % efficiency it follows that if we step down the voltage by ten times we also can step-up the current by ten times so, simplistically: -

\$V_{in} \cdot I_{in} = V_{out} \cdot I_{out}\$

The transformer is at the heart of an off-line switching power supply but instead of the transformer being fed with raw AC at 60Hz or 50Hz it is fed at a much higher frequency and this means a much smaller transformer can be used.

So the incoming 110V AC is rectified and smoothed (not to 110V DC) but to a voltage more like 154 volts DC. This feeds a power oscillator (that might run at 100kHz) and this drives the transformer primary.

On the secondary (the low voltage side) there is another rectifier and more smoothing capacitors. There may also be a feedback system that keeps the output closely regulated. Here's a typical block diagram: -


(source: philpem.me.uk)  


  So I should be able to get by with a smaller gauge wire going to the
  input (at high voltage) of my switching power supply while having to
  use a heavier gauge wire for the output (at lower voltage).


Correct!