Gain of common-emitter BJT amplifier with bypass capacitor
There are several gains associated with voltage amplifiers.  Consider the following model 



In this model, the gain \$A_{VO} \$ is the open circuit voltage gain of the amplifier which, in your circuit, is given by \$R_C/r_e\$

But note that the output resistance of the amplifier (which is about \$R_C\$ in your circuit) forms a voltage divider with the load \$R_L\$.

So, the loaded voltage gain is:

\$A_{V} = A_{VO} \dfrac{R_L}{R_{out}+ R_L}\$

But, note that the input voltage \$V\$ is less than the source voltage due to the voltage division between the source resistance and the input resistance of your amplifier.  Thus, the loaded gain with respect to the source is:

\$A_{VS} = A_V \dfrac{R_{in}}{R_S + R_{in}} = A_{VO} \dfrac{R_L}{R_{out}+ R_L} \dfrac{R_{in}}{R_S + R_{in}}\$

So, you cannot expect to measure anything close to the open circuit gain \$R_C/r_e \$