In an op-amp, why does Vn approach Vp until the difference is delta V / A? Why doesn't Vn just equal Vp?
Because the opamp has finite gain.  Note that in your equation when A is infinity, Vp and Vn are equal.

Since the opamp has finite gain, it needs some small difference between Vp and Vn to produce a non-zero result.  For example, let's say the opamp gain is 100,000 and is producing 7 V out.  That means it must see a difference of (7 V)/100,000 = 70 &micro;V to produce that 7 V out.

We can usually consider the two inputs to be the same voltage and ignore a small difference like 70 &micro;V.  Usually the input offset voltage swamps that anyway.  There isn't much point considering 70 &micro;V difference when the input offset voltage is 2 mV.  That means that the actual voltage difference between Vp and Vn can be up to 2 mV off from what it should be ideally.  Basically, the input offset voltage is the error inside the opamp when interpreting the difference between its inputs.  In this example, the additional offset of 70 &micro;V to make the opamp produce its actual output is only 3.5% of the ambiguity in the difference between the inputs anyway.