How does a common emitter amplifier work (intuitively)
Here is a very simplistic intuitive explanation of this most widespread transistor circuit:

Think of the transistor as a "voltage-controlled resistor" (with the proviso that it is non-linear)... something like a rheostat whose slider is driven by the input voltage. The input of this device (where voltage is applied) is the base-emitter junction, and its output (where resistance appears) is the collector-emitter junction. Thus, we may consider the "complex" 3-terminal transistor as composed of two simpler 2-terminal elements. Now it remains to decide where to apply the input voltage and where to take the output voltage from.

If we do not know about the principle of the common ground in circuitry, we are happy people... and we simply apply the input voltage to the base-emitter junction ("+" to the base, "-" to the emitter). To take the output voltage, we make a "voltage divider" by connecting an additional (collector) resistor in series to the transistor output. We supply this network and take the voltage drop across one of them as the output voltage.

However, we soon learn that (for a number of reasons) in circuitry devices are connected by one of their terminals to a common reference point called “ground”... and we begin looking for a solution to the problem.

We are inventive enough to present the input (differential) voltage as a difference between two input single-ended (grounded) voltages applied to the base and emitter... and thus we solve the problem with the common ground. Now we can change the base voltage as an input while keeping the emitter voltage constant... or change the emitter voltage as input by keeping the base voltage constant... we can even change both... but this is another story... With respect to the voltage change, the constant voltage behaves as an (AC) ground; hence the name “common emitter” and “common base”.

In the common emitter stage we keep the emitter voltage constant, particularly zero (grounded emitter) and change the base voltage as an input.