There is no power increase. You are simply using the energy storage of the cpacitor and the ability for the transformer/rectifier/diode to provide high peak currents to boost the average DC voltage available.
In the example above, lets say that the diodes conduct 20% of the time, or about 2 ms. The ripple voltage is about 2 V. So the capacitor loses 2 V in the 80% period and must be charged up during the 2ms time period. Since C = q/v, for a 40,000 uF capacitor to increase by 2 V, you need .08 coulombs. SInce an amp is one columb per second, you need to dump .08 coulombs in 2 ms, or 40 A of peak current during the conduction time.
So, no magic involved, assuimg all the components can handle the peak currents, the capacitor is kept charged at a DC voltage higher than the RMS value of the AC voltage.