What is the point of using an inverting amplifier in a circuit? This may be a simple question to most of you, but I'm having trouble understanding the concept behind the use of an inverting amplifier.
What is the purpose of using an op amp in an inverting amplifier if all of the current flows through the feedback resistor. In my textbook it states that the potential on the inverting input is equal to the non-inverting input which is Earthed, which produces a differential input of 0V. How is this in any way useful for the function of the circuit?
Also, how is the potential at $V_2$ 0? If current flows from $V_{in}$ to $V_{out}$ then how is the potential at that point 0? My textbook also states the currents $I_{in}$ and $I_F$ are in opposite directions seeing as $I_F=-I_{in}$. If this is the case I can see why the potential at $V_2$ is 0, but why are the currents in the opposite directions?

 A: One of the properties ("golden rules") of an ideal operational amplifier is it drives the $V_+$ input voltage to the input voltage $V_-$ (when operated closed loop).  So, in your case, $V_-$ is grounded so the op-amp will function to pull $V_+$ to ground.
One advantage/use of an inverting op-amp is to amplify a small signal ($V_{in}$ in your case) since the gain is,
$$\frac{V_{out}}{V_{in}}\approx-\frac{R_f}{R_{in}}$$
So, in your case, if $V_{in} = 1.0V$, $R_f=100kΩ$, and $R_{in}=10kΩ$ then,
$$V_{out} \approx -10.0V$$
Since the op-amp is a very high impedance device $\approx 0$ current will flow into the inputs, so we can say,
$$i_{in}+i_{f}=0$$
And,
$$i_{in}=\frac{V_{in}}{R_{in}}=\frac{1.0V}{10kΩ}=0.1mA $$
$$i_{if}=\frac{V_{out}}{R_{f}}=\frac{-10.0V}{100kΩ}=-0.1mA $$
The actual currents flowing in our example are thus,


The assumed current arrows are in opposite directions (top figure), but because of the inversion giving us a negative sign, you will see that the actual currents are as shown in second figure.  No current (of significance) flows into the op-amp inputs.  The output however, is sinking 0.1mA.
UPDATE showing simplification.
Solve this simple circuit for the voltage $V_X$ and see if it helps.

