Input impedance of grounded emitter amplifier with DC feedback
It's mostly about how the collector voltage changes with respect to a base voltage change.

To start, it's assumed that the input capacitance's impedance is negligible at the frequency under discussion and can be neglected. Next, you know that the voltage gain is near \$A_v=\frac{R_{_\text{C}}}{r_e}\approx 320\$ (given your own figures here.)

Since the collector voltage moves in the opposite direction to the base voltage, and does so by that gain figure, then the change in voltage across the resistor will be \$v_{_{\text{R}_1}}=-A_v\cdot v_i - v_i=-\left(A_v+1\right)v_i\$. Therefore, \$i_{_{\text{R}_1}}=\frac{v_{_{\text{R}_1}}}{R_1}=-\left(A_v+1\right)\frac{v_i}{R_1}\$.

Normally, you'd expect \$\mid\: i_{_{\text{R}_1}}\mid=\frac{\mid v_i\mid }{R_1}\$. But as you can see, it is \$\mid-\left(A_v+1\right)\mid=A_v+1\$ times larger. So the resistance of \$R_1\$, as seen by the input, appears to be about 321 times smaller because of the direction and magnitude of the collector voltage change with respect to the base voltage change.