Why the input resistance of a common emitter amplifier is like this?
Draw this small-signal equivalent circuit: 

\$R_{IN} = \frac{V_X}{I_B}\$ 







simulate this circuit &ndash; Schematic created using CircuitLab

And we can see that \$R_{IN} = \frac{V_X}{I_B}\$ 

$$V_X = I_B\cdot r_\pi + I_ER_E = I_B\cdot r_\pi + (I_B + I_C )R_E = I_B\cdot r_\pi + (I_B + h_{FE}I_C )R_E
  $$
$$=I_B\cdot r_\pi +I_B(h_{FE}+1)R_E$$

Therefore 

\$R_{IN} = \frac{V_X}{I_B} =r_\pi + (h_{FE}+1)R_E\$

Or simply think about emitter current

\$I_E = I_B + I_C = I_B + \beta I_C = I_B(\beta+1) \$

Try read this 
http://www.ittc.ku.edu/~jstiles/412/handouts/5.6%20Small%20Signal%20Operation%20and%20Models/section%205_6%20%20Small%20Signal%20Operation%20and%20Models%20lecture.pdf