Why can't I light a LED with a 1.5 V battery?
LEDs don't work like ordinary (incandescence) light bulbs.
Main differences (a bit simplified for very beginners):
They have a polarity, hence they must be powered using DC respecting that polarity. Reverse the polarity and they won't work. You may also damage them if you apply more than ~4V-5V in the reverse direction (these are safe values; the exact maximum tolerable value depends on the specific device).
Light emission begins only if a certain voltage is reached (threshold voltage), under that voltage the emission is negligible. Therefore, if you have a battery whose voltage is below the threshold voltage of the LED, you are out of luck, unless you use a more complicated circuit (e.g. a Joule thief or a boost DC-DC converter) to power the LED.
After the threshold voltage is reached, any very slight increase in voltage makes the LED conduct heavily, i.e. absorb a huge current. Hence you need a resistor in series to limit that current to a safe limit. There are other questions/answers on this site explaining how to calculate the value of the limiting resistor.
Once conducting, the light intensity emitted is roughly proportional to the current (not voltage) that flows in the diode (so you get a brighter LED if you decrease the value of the limiting resistor). This up to the maximum current limit of the LED. After that limit has been reached, the device goes POOF!
You also ask why all this happens, but the answer is fairly complex, since it depends on the physical structure of the semiconductor crystal inside the diode. The physical explanation lies in quantum mechanics and solid-state physics, really tough subjects.
The Wikipedia article on LEDs only scrapes the surface of the internal workings of the LEDs and still is fairly complex.
I see Lorenzo has already answered your question directly (+1). Here is what you can do to light your LED and see what you've got.
LEDs are diodes, so conduct only in one direction. Unlike a ordinary light bulb, orientation matters. If the LED doesn't light one way, flip it around and try again.
To safely experiment with pretty much any LED, use a 5 V supply with at least 180 Ω in series. Using a higher resistance works, but will light the LED more dimly. Even with 1 kΩ in series, you will still be able to see any visible-light LED light up indoors.
The reason to use a 5 V supply is to limit the reverse voltage across the LED when connected backwards. Most LEDs can stand at least 5 V reverse across them.
A visible-light LED will drop a minimum of 1.8 V. That leaves (5 V)-(1.8 V) = 3.2 V across the resistor. Just about any LED can handle 20 mA forward current. By Ohm's law, (3.2 V)/(20 mA) = 160 Ω. I said 180 Ω minimum for a little margin and because that is a common value.
LED forward voltage is dependent on color. Common green LEDs drop about 2.1 V, for example. "White" LEDs are usually really UV LEDs with phosphors that re-emits in the visible spectrum. Those can drop around 3.5 V.
With a 200 Ω resistor and a 3.5 V LED, you get (1.5 V)/(200 Ω) = 7.5 mA. Such a LED will still light quite visibly with 7.5 mA thru it, even if it could have handled 20 mA or more.
Once you get your LED to light, you can measure its forward voltage, then adjust the resistor to allow the maximum current with that forward voltage. Assume the maximum is 20 mA unless you have a datasheet and it says otherwise.
Physics explanation
Light bulbs
Incandescent light isn't really a light source so much as a heating element. Any current through a wire heats it up a bit (Joule's Law of Heating); once the wire is above room temperature it emits net energy via black-body radiation. The rate at which this energy is emitted depends on the fourth power of temperature, i.e. the higher the temperature the brighter†. And the more current (or equivalently‡ more voltage), the higher the temperature of the wire.
The fundamental physical process behind blackbody light emission is this: the atoms in a warm piece of material are shaken around by thermal motion. This motion is completely chaotic, so even if the average energy per atom is rather low, every once in a while an atom at the surface will get a push from multiple neighbours and thus gather enough energy that it can emit a visible photon (at least \$2.6\times10^{-19}\$ Joules). But far more often, it will only have enough energy to emit an invisible infrared photon.
LEDs
By contrast, LEDs pump the atoms directly to the energy that's required for emitting visible light. They do this by cleverly exploiting the band gap of a semiconductor. That's a quantum-mechanical feature of crystals like silicon, which “forbids” electrons from having energies in a certain range. You then take one piece of semiconductor which has been doped so the conductance electrons are all above the band gap, and one where they're all below the bandgap. Then, when a current flows across the junction, each electron loses just the right amount of energy to excite an atom to produce a photon with the right energy to be visible – again, for red light, this is about \$2.6\ldots3.2\times10^{-19}\$ Joules.
Only... why would the electrons continue to go over the junction? After an electron has crossed the junction, it won't be inclined to climb across the band gap again; that costs energy which the electron doesn't have. ...Unless you give it the energy from an external source: each volt that you apply to a circuit can supply an electron with an energy of \$1.6\times10^{-19}\:\mathrm{J}\$, a quantity that physicists call just an electron volt. So when you apply a voltage of \$U\$ to an LED whose band gap has an energy of \$U\times1\:\mathrm{eV}\$, you can keep up a current. This voltage isn't really dependent on how much current actually passes through the LED, therefore the brightness can't effectively be regulated by tweaking the voltage – you need to regulate the current instead. And if the voltage falls below the band gap, the current will just cease entirely, because the conductance electrons just won't go into the n-doped domain at all anymore.
†That's a bit too simplistic: Stefan-Boltzmann describes the intensity integrated over the entire electromagnetic spectrum. Only a narrow band of that is actually visible (that's the reason incandescent light is so much less efficient than LEDs). Since the wavelength of peak intensity also depends on the temperature, brightness is in fact related not just to \$T^4\$ but to a more complicated relation, but still: higher temperatures always correspond to brighter light.
‡Similarly, Ohm's law isn't completely correct here because the resistivity depends on temperature. But the qualitative dependency higher voltage ⇒ higher electrical power still holds true.