Understanding simple circuit with two transistors and one led

Question

I'm trying to understand the following circuit:

I expected it to work like this: Initially the LED is dark. When S1 is pressed, Q2 is "opened" and powers Q1's base, so Q1 is "opened" and current can flow to the LED. In addition, because Q1's emmitter is connected to Q2's base, I expected that both transistors will stay "open" and the LED stays lit up even when S1 is released.

But it doesn't work like that. When S1 is pressed, the LED is lit up, but when S1 is released, the LED goes dark again. Why is that?

EDIT: Added D2:

Answer 1

If S1 is N.O. (normally open):

When you press S1, R1 is connected to 9V, so the transistors are shorted out and your circuit acts like a simple LED+resistor config.

When you release S1, no current flows through Q2's base, therefore it allows no current to flow through Q1's base and Q1 does not conduct, so current does not flow at the switch nor the transistors, so the LED stays off.

p.s.: apparently you use "open" as conducting and "closed" as not conducting, this might generate misunderstandings as in electronics "open" circuit is a circuit that is not conducting, e.g.:

Answer 2

If you want to have the led to stay lit even after the switch is opened you must use this circuit...

Answer 3

^{simulate this circuit – Schematic created using CircuitLab}

Figure 1. Voltages through the circuit.

You are forgetting that each base-emitter junction will drop about 0.7 V.

Looking at Figure 1 you can see that when SW1 is closed its base will be at 9 V but it's collector will be at 8.3 V. Similarly another 0.7 V will be dropped across Q2's base-emitter junction so there will be only 7.6 V at it's emitter.

Obviously you can't use that voltage to feed back into Q1 to keep the whole circuit alive.