Precision half wave rectifier
You are imagining it wrong. The feedback loop reduces the gain at he same time as signal is amplified, not after that.

Op-amp can be modeled as a feedback control system:



The output signal is amplified by \$\frac{A}{A+1}\$, never more. I agree that with ideal step input, it feels like a chicken-or-egg problem. But in reality, there is always finite slew rate that will allow the signal to propagate from output to input before it gets to \$A\cdot V_{IN}\$.

As a side node, in reality the rectifier circuit needs two feedback paths, to prevent the op-amp output from saturating (because unlike comparators, for op-amps it takes ages to get out of saturation).

Edit: The diode will not make much difference, because it is inside the feedback loop and will simply cause the op-amp output to rise by its forward voltage drop, maintaining its both inputs at the same voltage.



Anode-cathode voltage \$V_{AC}\$, even though it is not constant, will make very little change because the output voltage is \$V_{OUT}=\frac{A\cdot V_{IN}-V_{AC}}{1+A}\$ which is almost equal to \$V_{IN}\$.