What is wrong with my nested loops in Python?

Question

How do I make nested loops in Python (version 3.0)?

I am trying to get the following loops to show me the products of two numbers:

def PrintProductsBelowNumber(number):
    number1 = 1
    number2 = 1
    while number1 <= number:
        while number2 <= number:
            print(number1, "*", number2, "=", number1 * number2)
            number2 += 1
        number1 += 1

PrintProductsBelowNumber(2)

As a result I get:

1 * 1 = 1
1 * 2 = 2

So it seems the outer loop over number1 does not run.

How do I get the loop over number1 to run, and thus obtain:

1 * 1 = 1
1 * 2 = 2
2 * 1 = 2
2 * 2 = 4

Answer 1

Because you aren't setting number2 back to 1 after the inner loop completes the first time. number1 then increments, but since number2 is still too high the inner loop doesn't run again.

def PrintProductsBelowNumber(number):
    number1 = 1
    while number1 <= number:
        number2 = 1
        while number2 <= number:
            print(number1, "*", number2, "=", number1 * number2)
            number2 += 1
        number1 += 1

PrintProductsBelowNumber(2)

EDIT: Adam's solution is much better in general, but this is to show why your's wasn't working the way you thought it should in the first place.

Answer 2

number2 only gets initialized once, you need to re-initialize it for each iteration of the inner loop. However, this code is very C-like and not very Pythonic. The better way to do it would be to use the for number in range(n) construct:

def PrintProductsBelowNumber(number):
    for number1 in range(1, number+1):
        for number2 in range(1, number+1):
            print(number1, "*", number2, "=", number1 * number2)

Answer 3

You could modify Adam's solution with a list comprehension:

def PrintProductsBelowNumber(number):

    results = [(i, j, i * j) for i in range(1, number + 1) 
                             for j in range(1, number + 1)]

    for number1, number2, result in results:
        print(number1, "*", number2, "=", result)

or some variation thereof.