Why is feedback required in op-amp circuits?
You already know that an opamp has very high open-loop amplification, typically 100 000 times. Let's look at the most simple feedback situation:  

  

The opamp will amplify the difference between \$V_+\$ and \$V_-\$:  

\$ V_{OUT} = 100 000 \times (V_+ - V_-) \$  

Now \$V_+ = V_{IN}\$ and \$V- = V_{OUT}\$, then

\$ V_{OUT} = 100 000 \times (V_{IN} - V_{OUT}) \$  

or, rearranging:  

\$ V_{OUT} = \dfrac{100 000}{100 000 + 1} \times V_{IN}\$  

That's as good as 

\$ V_{OUT} = V_{IN}\$

This is a voltage follower, a \$\times\$1 amplifier, which is mostly used to get a high input impedance and a low output impedance.

The feedback reduces the very high open-loop amplification to \$\times\$1. Note that the high amplification is needed to get \$ V_{OUT}\$ as close as possible to \$V_{IN}\$.

edit
Now by using just a fraction of the output voltage in the feedback we can control the amplification.  

  

Again

\$ V_{OUT} = 100 000 \times (V_+ - V_-) \$,   

but now \$V_+ = V_{IN}\$ and \$V- = \dfrac{R1}{R1+R2} \times V_{OUT}\$, then

\$ V_{OUT} = 100 000 \times (V_{IN} - \dfrac{R1}{R1+R2} \times V_{OUT}) \$

Or:

\$ V_{OUT} = \dfrac{100000 \times V_{IN}}{\dfrac{R1}{R1+R2} \times 100000 + 1}  \$

The term "1" can be ignored, so that  

\$ V_{OUT} = \dfrac{R1+R2}{R1} \times V_{IN}  \$  

Notice that in both the voltage follower and this non-inverting amplifier the actual amplification factor of the opamp cancels provided it is high enough (>> 1).