Given an Op Amp where the output voltage is fed back into the negative terminal, why is the gain one?
The gain of the op amp is very high, typically 100k to 1M or more.  The gain of the circuit in this case is very close to 1.

For now, let's imagine a nice ideal opamp with 0 offset voltage, infinite gain, etc.  Think of what would happen if the gain weren't 1.  Let's say you drive Vin to 3.0 V.  The opamp will then drive Vout to also be 3.0 V.

Now imagine what would happen if it didn't.  Let's say the opamp only drove Vout to 2.9 V.  That means the positive input of the opamp is 3.0 V and the negative input 2.9 V, for a total differential input signal of 100 mV.  Multiply that by the gain of the opamp, and that's what it will try to drive its output to.  Obviously it will try to drive Vout as high as it can in this case.  However, eventually Vout gets to 3.0 V, the differential input to the opamp is then 0 and it no longer tries to drive Vout high.  If Vout went a tiny bit higher, the differential input to the opamp would be a little negative, which would make it try to pull Vout low.  Put another way, the opamp will try to pull Vout down if it is above Vin, and up if it is below Vin.  Eventually, assuming nothing is externally holding Vout fixed, Vout converges to be the same as Vin.

Added:

This can be described mathematically quite easily too.  What a opamp does is basically:

&nbsp;&nbsp;Vo = G(Vp - Vn)

Where Vp is the voltage at the positive input, Vn the voltage at the negative input, Vo the output voltage, and G the gain of the opamp.  As I said above, G is usually a large number, like 100k to 1M or more.

From how the circuit is wired, we get:

&nbsp;&nbsp;Vn = Vo
&nbsp;&nbsp;Vp = Vin

Substituting these into the first equation, we get:

&nbsp;&nbsp;Vo = G(Vin - Vo)

Solving for Vo as a function of Vin yields:

&nbsp;&nbsp;Vo = (G / (1 + G)) Vin

Note that G / (1 + G) is basically 1 for large values of G.  Even if G is only 100k, then the overall gain of this circuit is 0.99999, which is 1 for the vast majority of practical purposes.