This question would only apply to the "old fashioned" incandescent light bulbs an not fluorescent, compact fluorescent, or LED bulbs. The incandescent bulb works by having an electric current pass through a very thin wire. The current passing through the wire encounters resistance due to the chemical composition of the wire and the diameter of the wire (thinner the wire, the greater the resistance). One effect of resistance is to convert the electrical energy in the current to heat energy.
The light bulb is designed in such a way as to create enough heat energy to cause the wire, or filament, of the bulb go get hot enough to glow thus producing light. However, as one heats and cools the filament over time the metal undergoes two changes.
The heat causes the metal to vaporize slightly; the loss of metal over time eventually causes the wire to thin so much that the heat of resistance can rise above the melting point of the metal in the wire causing it to break. Another change is that the heating and cooling causes an "annealing" process-- that is that it gets brittle over time.
When the bulb is turned on there is a sudden expansion of the wire filament due to the increase in temperature. If the wire has become very brittle or very thin the sudden expansion can cause the wire to break. In either case the broken wire creates an 'open circuit" and the bulb "burns out".