Any better way to solve Project Euler Problem #5?

Here's my attempt at Project Euler Problem #5, which is looking quite clumsy when seen the first time. Is there any better way to solve this? Or any built-in library that already does some part of the problem?

``
'''
Problem 5:
2520 is the smallest number that can be divided by each of the numbers from 1 to 10

What is the smallest number, that is evenly divisible by each of the numbers from
1 to 20?
'''

from collections import defaultdict

def smallest_number_divisible(start, end):
'''
Function that calculates LCM of all the numbers from start to end
It breaks each number into it's prime factorization,
simultaneously keeping track of highest power of each prime number
'''
# Dictionary to store highest power of each prime number.
prime_power = defaultdict(int)

for num in xrange(start, end + 1):
# Prime number generator to generate all primes till num
prime_gen = (each_num for each_num in range(2, num + 1) if is_prime(each_num))

# Iterate over all the prime numbers
for prime in prime_gen:
# initially quotient should be 0 for this prime numbers
# Will be increased, if the num is divisible by the current prime
quotient = 0

# Iterate until num is still divisible by current prime
while num % prime == 0:
num = num / prime
quotient += 1

# If quotient of this priime in dictionary is less than new quotient,
# update dictionary with new quotient
if prime_power[prime] < quotient:
prime_power[prime] = quotient

# Time to get product of each prime raised to corresponding power
product = 1

# Get each prime number with power
for prime, power in prime_power.iteritems():
product *= prime ** power

return product

def is_prime(num):
'''
Function that takes a number` and checks whether it's prime

You are recomputing the list of prime numbers for each iteration. Do it just once and reuse it. There are also better ways of computing them other than trial division, the sieve of Eratosthenes is very simple yet effective, and will get you a long way in Project Euler. Also, the factors of n are all smaller than n**0.5, so you can break out earlier from your checks.

So add this before the num for loop:

prime_numbers = list_of_primes(int(end**0.5))

And replace prime_gen with :

prime_gen =(each_prime for each_prime in prime_numbers if each_prime <= int(num**0.5))

The list_of_primes function could be like this using trial division :

def list_of_primes(n)
    """Returns a list of all the primes below n"""
    ret = []
    for j in xrange(2, n + 1) :
        for k in xrange(2, int(j**0.5) + 1) :
            if j % k == 0 :
                break
        else :
            ret.append(j)
    return ret

But you are better off with a very basic sieve of Erathostenes:

def list_of_primes(n) :
    sieve = [True for j in xrange(2, n + 1)]
    for j in xrange(2, int(sqrt(n)) + 1) :
        i = j - 2
        if sieve[j - 2]:
            for k in range(j * j, n + 1, j) :
                sieve[k - 2] = False
    return [j for j in xrange(2, n + 1) if sieve[j - 2]]

There is an alternative, better for most cases, definitely for Project Euler #5, way of going about calculating the least common multiple, using the greatest common divisor and Euclid's algorithm:

def gcd(a, b) :
    while b != 0 :
        a, b = b, a % b
    return a

def lcm(a, b) :
    return a // gcd(a, b) * b

reduce(lcm, xrange(start, end + 1))

On my netbook this gets Project Euler's correct result lightning fast:

In [2]: %timeit reduce(lcm, xrange(1, 21))
10000 loops, best of 3: 69.4 us per loop

StackExchange Code Review Q#21367, answer score: 7