Reason of saturation region in MOSFET I can't wrap my head around the circumstance that, somewhen when I increase $ V_{DS} $ the current $ I_{DS} $ will remain constantly. Why is that?

https://www.electronics-tutorials.ws/amplifier/mosfet-amplifier.html
 A: For concreteness, consider an n-channel enhancement mode MOSFET whose source terminal is connected to the body.  If $V_{DS}=0$ and $V_{GS}>V_{th}$, a conducting channel is opened between the drain and source.  When $V_{DS}$ is increased, current flows across the channel.
However, increasing $V_{DS}$ has another effect - it changes the charge distribution on the gate.  When the source voltage is higher than the drain voltage, the positive charge which was initially (more or less) evenly distributed across the gate gets pushed toward the drain side.  This has the effect of making the conducting channel asymmetric - wider near the drain and narrower near the source.
If $V_{DS}$ exceeds $V_{GS}-V_{th}$, then the channel will be closed near the source region.  This is referred to the pinch-off regime.  Past this point, increasing $V_{DS}$ causes a proportional increase in resistance because the channel closes off even further away from the source region.  As a result, the current does not appreciably increase as $V_{DS}$ increases.  This is why the current saturates.
The current does increase with changes in $V_{GS}$, however, and the increase can be quite dramatic.  That's why amplifiers operate in this regime - a signal on the $V_{GS}$ terminal produces a current signal $I_{DS}$.
