What you have is essentially this:
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simulate this circuit– Schematic created using CircuitLab
If we assume about \$700\:\text{mV}\$ for the \$V_{_\text{BE}}\$ junction of the NPN, tentatively, and similarly assume about \$2\:\text{V}\$ for a red LED, also tentatively, then this works out to about \$\frac{0\:\text{V}-700\:\text{mV}-2\:\text{V}-\left(-5\:\text{V}\right)}{1\:\text{k}\Omega}=2.3\:\text{mA}\$ of emitter current in the NPN bipolar.
While that may look fine in the simulator, it may not be enough current to make the LED visible, in practice.
If you want to keep your design (I wouldn't, but that's a separate issue for later), then perhaps you could modify both resistors you are using to a values of about \$120 \:\Omega\$.
It's not a good design. But if you try out those suggested replacement values, you may see better results in your existing circuit.
