Output impedance of current series feedback BJT amplifier?
The "gain" formula as given by you is the transconductance of a transistor in common-emitter configuration WITH feedback. 

1.) Without feedback (RE=0) this transconductance is simply 
$$gm = \frac{hfe}{hie}$$

2.) With feedback we have 
                            $$A= \frac{gm}{1+gm*R_E}= \frac{hfe}{hie+R_E}$$

From this, we can derive the loop gain Aloop=-gm*Re.
Note that gm=hfe/hie.

3.) Based on the loop gain expression (and knowing that the output resistance will be increased using feedback) we can expect (approximation):

$$Z_{out}= Z_o*(1+gm*R_E)=r_{ce}*(1+gm*R_E)$$

4.) Comment: The above expression for Zout is an approximation because it is derived under the assumption of IDEAL voltage feedback. However, the feedback voltage is provided across RE (not an ideal voltage source). More than that, also the influence of the input circuitry was neglected. However, we should not forget, that during practical operation of this circuit the large value of Zout is paralleled with the collector resitor Rc (which is much smaller that Zout). Hence, the approximation seems to be acceptable.