Why does a higher wattage incandescent light bulb have a lower resistance value than a lower wattage incandescent light bulb? I am an electrician and know through experience that resistance in an electrical circuit causes heat. An incandescent light bulb's light is a by-product of heat, so why does a 100w bulb have a lower resistance than a 25w bulb? It seems counter intuitive to me. Please help me understand.
 A: You probably know that $P=IV$ (power is current times voltage), right?  And Ohm's Law, $I=\frac{V}{R}$.
If you substitute $\frac{V}{R}$ for $I$ in the power equation, you will find your answer.


What confuses me, is that apparently for an incandescent light bulb, less resistance=more heat?

That's what the math says:   $P=\frac{V^2}{R}$.  Power (in Watts) is equal to the square of the Voltage divided by the resistance.
What's happening is, the national power grid does a very good impersonation of a constant voltage source.  No matter what you connect to your power line, if the voltage is 120V when you're not drawing any current, it still will be 120V when you're drawing tens or hundreds of amperes.
Ohm's law says that if the voltage is constant, then there will be more current for a smaller resistance.  The power law says that if the voltage is constant, then more current means more power.
Therefore, if the voltage is constant, there will be more power consumed by a smaller resistor.


...does this mean that a heating element, such as in a dryer or oven, must have less resistance than the wires supplying the power?

Good question.  No.  It doesn't mean that.
The water heater and the wires form a circuit called a voltage divider.  Physically, there are three resistors; One conductor from the panel to the heating element, the heating element itself, and the other wire back to the panel.
The math is easier though if we just consider two resistors:

The sum of the voltages dropped by each resistor in the loop, $V_1+V_2$, must add up to the supply voltage.  (That means I was lying a little bit when I said that the voltage supplied to your water heater was constant.  It's not.  When current flows in the circuit, the resistance of the wires steals some of the voltage.)  Meanwhile, the current flowing around the loop, $I$, must be the same everywhere because there's no place else for it to go.
If we say that $R_1$ represents the resistance of the wires, and $R_2$ represents the resistance of the heating element, you can see that $V_1$ will proportionately less that $V_2$ if $R_1$ is less than $R_2$.
That's important because Ohm's Law and the power law apply separately to each resistor:   $P_1=\frac{V_1^2}{R_1}$, and $P_2=\frac{V_2^2}{R_2}$.
Another law you can derive is $P=I^2R$, or in this case, $P_1=I^2R_1$ and $P_2=I^2R_2$.  So, we actually want the resistance of the wires to be much less than the resistance of the heating element because when they are wired in series, the smaller resistor will dissipate proportionately less power.
I don't have time to go into more detail, but here's a link to a tutorial/experiment that you can work through if you want to get a better feel for it:
http://www.ibiblio.org/kuphaldt/electricCircuits/Exper/EXP_3.html#xtocid113033
A: Steve, hopefully you'll get this.  Comment if you do!  
To answer your first question:  A 100W light bulb has lower resistance because as long as the light bulb resistance is higher than the wire resistance, you can take advantage of the equation P=I squared R.  Just decrease the resistance a little and the current is increased. Since the increased current is squared you get a lot more power.  The main thing to remember is you can't drop the resistance too much or the wires will become the load and they will heat up more.  You can change the load (or transfer a part of the load) to the wires by either making the wires thinner, longer, or replacing the light bulb with wire.
To your other question.  The heating elements must have a a higher resistance than the wires.  If the heating elements had a lower resistance then the wires would heat up.  
To summarize, you want to have your load have a high enough resistance where it becomes the load.  Then you want to take advantage of getting higher power by reducing the resistance (but not lower than the wires).  I'm sure they have reasons why they pick certain resistance levels.
A: A non-math way to think of it is that higher resistance means less electricity gets through. A higher watt bulb allows more electricity through, therefore the resistance must be lower.
A: I understand your question perfectly. 
Lets leave out all the fancy maths. 
You are correct in being slightly confused. Logically, a higher resistance causes more heat but we must also remember that a higher resistance causes less current. 
It is the product of the current and the resistance that generates the heat (not only the resistance). 
So although a 100W lightbulb has a lower resistance, it will have a much larger current due to the low resistance. The product of the current and resistance will be high!
Hope that helps. 
A: Generally it is the resistance of the bulbs thats actually constant. The wattage rating on the bulbs is given by looking at the power (which can be heat or light) generated by connecting the bulb across the 220V mains (or whatever the value is in your country) in parallel. If the bulbs are connected in some other configuration, or across some other voltage, the power output would be different but the resistance would still be the same. 
The voltage is constant, and the formula for power that is most suitable to use here is $P=V^2/R$ as it depends on voltage and resistance, both of which we know to be constant, as we connect the bulb parallel to mains. The other and arguably more well known formula $P=I^2R$ depends on current, which is variable here in the parallel arrangement. 
In the first one, power decreases with resistance, which seems counterintuitive but makes sense if we consider that greater resistance means less current.
