Common collector effect of emitter resistor on output impedance
Inside the BJT transistor (and hidden from view) is a resistor called \$r_E\$. Sometimes it is called \$r_{\pi}\$ just to confuse everybody but I was brought up with \$r_E\$ so I'll refer to it as that. I'll refer to the external emitter resistor as \$R_E\$. This is what \$r_E\$ looks like in the bigger picture: -



The value of \$r_E\$ is inversely proportional to the current flowing from the emitter (largely the current flowing into the collector) and, proportional to a value called \$V_T\$ and \$V_T\$ is 26 mV at room temperature.

So, \$r_E\$ is about 26 ohm per mA of emitter current. If you want more evidence of this see the extract I took from here regarding what \$V_T\$ is. I've put this under the main body of my answer at the bottom of the page.

In a common-collector amplifier, the "imaginery emitter" would follow the base signal but the real emitter voltage would be the imaginery emitter voltage minus the loading effect of \$R_E\$ on \$r_E\$. In other words these two resistors form a potential divider on the "imaginary emitter".

And this is why the output from a common collector (emitter follower) BJT circuit contains both resistor values in the formula. It's also why the output impedance of a common emitter is loosely referred to as \$r_E\$ but is more accurately referred to as $$r_E || R_E$$.



Where \$V_T\$ comes from: -