You have to keep in mind that when designing gear reductions, the torque is inversely proportional to the speed. As power (watts) = volts x amps in electrical design, the same power is transmitted at 72V 30A as 36V 60A. Similarly for mechanical systems, power (watts) = torque (Nm) x velocity (rad/s). The same mechanical power is transmitted in a shaft at 1000 RPM, 10 Nm and 100 RPM, 100Nm.
So the actual torque at your motor is not going to be all that much, since it's spinning at several thousand RPM. For a quick and dirty calculation, say your motor is 85% efficient and it's drawing 1200W of electrical power. So .85 * 1200W = 1020 W of mechanical power. If the motor is spinning at 4000 RPM, to use SI units you must first convert RPM to radians/sec:
4000 revolution/min = 4000 * (2pi) = 8000pi radian/min
8000pi radian/min = 8000pi / 60 = 133.33pi radian/sec or 418.8 rad/s
Now Power = torque * velocity, so torque = power / velocity. 1020 W / 418.8 rad/s = 2.43 Nm. That's it. No problem for set screws.
Now if you have, say, a 16:1 reduction, then your final output is going to be 418.8 / 16 = 26.1 rad/s, and
2.43 Nm * 16 = 38.9 Nm. Kind of a problem for a single set screw.
Just to check, final output power = 38.9 Nm * 26.1 rad/s = 1014 W. I chopped off some decimal points, but it's about the same, so we're good.
I hope this helps some. Let me know if anything is unclear and I'll try to explain it better.