current - Why so many pins for the MOSFET's drain?

The FDC855N comes in a 6-pin package, 4 of which are connected to the drain, and only 1 to the source. Why this difference? The source sees the same current as the drain, doesn't it?

Answer

That's not for the high current, it's for heat management.

The single source pin can handle the current, and so would a single drain pin. Schematically a MOSFET is often drawn symmetrically, because this way it's easier to show the asymmetry in the channel's conductivity.

But discrete MOSFETs aren't constructed that way. More like this:

It will probably be packaged upside down, with the bulk of the drain connected to the lead frame which directly connects to the 4 pins. Gate and source will be bonded to their pins.
The bulk of the MOSFET will dissipate the most heat, and because its direct contact with the pins the heat can be drained through the pins, it's a path with low thermal resistance. The drain may still be wire bonded as well, for proper electrical connection. But the bonding wire will pass much less of the heat.
Thermal resistance in conduction (to the PCB's copper) is much lower than that of convection (the way heat is exchanged with the air above the package). I found the following suggested pad layout for a Luxeon power LED. They claim it can easily achieve 7K/W.

In SMT power MOSFETs which will have to dissipate quite some heat it's advisable to have the drain pins on a larger copper plane, or allow the heat to dissipate through a series of (filled) vias, like for the Luxeon LED.