Answer: C

It would be impractical to find the product of the integers from 1 to 20. Since we want to know the greatest power of 4 that is a factor of that product, we only need to focus on the 4's (and the factors of 4) in each of those 20 integers.
First of all, all of the odd integers between 1 and 20 are irrelevant. No matter how many odd integers you multiply together, the result will never be divisible by 4, let alone a multiple of 4. That leaves us with the evens.
When dealing with factors and multiples, it's always a good idea to work with primes. So rather than considering the number of multiples of 4 in the remaining numbers, let's focus on 2's. Between 1 and 20, there are 5 numbers (2, 6, 10, 14, 18) that are divisible by 2, but not by four. That means that, if we multiplied those five numbers together, their prime factorization would contain a 2^5.
There are five other numbers to consider, listed here with the number of 2's in their prime factorization:
4: 2^2
8: 2^3
12: 2^2
16: 2^4
20: 2^2
Add up the first five 2's and the 2's in each of those numbers, and we have a result of 2^18. That means: If we multiplied all of those numbers together, the prime factorization of the result would contain 18 2's, and the number would be divisible by 2^18.
But we care about 4's, not 2's. 4 = 2^2, so:
4^n = 2^18
(2^2)^n = 2^18
2^2n = 2^18
2n = 18
n = 9
4^9 is a factor of the product of the integers 1 to 20, inclusive.