Why Does Vout Become Constant in a MOSFET Circuit at High Vin?

[act858]

Homework Statement

I would like to plot Vout vs Vin for the following circuit:

Homework Equations

Square law equations for MOSFET transistors.

The Attempt at a Solution

I'm calling the threshold voltage Vth.

My understanding of the circuit is this:
1. From 0 < Vin < Vth, the transistor will be off, and no current will be flowing though Rs, making Vout = 0.
2. Now Vout = Vin - Vth until the drain voltage is at Vin - Vth. This is because as Vin increases, the current through Rs increases proportional to Vin^2, which pushes the voltage at Vout up proportional to Vin^2.
3. When the drain equals Vin - Vth the transistor will enter triode mode. Here's where I get confused. In triode, the current should be linear in Vin. So applying the same logic as in 2, Vout should continue to follow Vin - Vth. But when I run the simulation in SPICE, it stops and becomes constant. The current also becomes constant. (Vth = 1 for this transistor)

My book also has this as the answer and gives the value of Vout to be the voltage division of Rs and Rf. I'm not sure how to arrive here, it seems like the current is no longer dependent on Vin at this point when it should be. Can someone explain what's going on, specifically why the current stops depending on Vin? If the current is constant then I can accept that the voltage will be constant. But then, what would be the intuitive way to see that Vout = (Vdd * Rs) / (Rs + Rf)? Here's my books answer for reference:

 

[rude man]

See attached for standard MOSFET equations including the Early effect. Stick to the "simple analytical model".