Olin is right. We shouldn't have to watch an 11 minute video to get to your question. So I didn't watch it, but I understand your question is about ripple.
The dashed line is the voltage from your rectifier. The solid line is the voltage across the smoothing capacitor. Every half cycle of your AC supply (8.3ms in the US, 10ms in Europe) the input voltage will go higher than the capacitor's value and the capacitor will be charged through the diodes. There's little resistance, so the current peak can be quite large. This charging stops when the input voltage drops below the capacitor's voltage, when the sine is at its peak. From then on the capacitor gets discharged by the load, which makes its voltage sag. The discharging stops when the input voltage goes higher than the capacitor's voltage again, where the sine intersects with the smoothed ripple.
How fast the voltage drops depends on the capacitor's size and the load. A higher load (more current drawn) will make the voltage drop faster, causing a larger ripple. On the other hand a larger capacitor will hold more energy, so that that same load will make the voltage drop slower.
"What is the simplest way to go from A/C to D/C without much ripple?"
Don't load your capacitor and you'll have near zero ripple. :-)
OK, the load will be a given, so there's just the capacitance you can play with. You'll want a reasonably low ripple voltage, but you don't have to go to extremes. A voltage regulator will suppress the remaining ripple quite well. Just make sure you respect the regulator's minimum input voltage.