To find the number of consecutive zeroes at the end of a factorial number, we need to determine the number of times the number is divisible by 10. Since 10 is a product of 2 and 5, we need to count the number of times 2 and 5 appear as factors in the factorial number.

Factorial of a number n is denoted as n! and is the product of all positive integers from 1 to n. In this case, we need to find the number of consecutive zeroes at the end of 72!.

Prime Factorization:
To find the number of times 2 and 5 appear as factors in 72!, we can break down the factorial number into its prime factors.

- Prime factorization of 72:
72 = 2^3 * 3^2

- Prime factorization of 10:
10 = 2 * 5

Counting Factors of 2 and 5:
Since we need to find the number of factors of 2 and 5, we count the number of times 2 and 5 appear in the prime factorization of 72!:

- Factors of 2:
The prime factorization of 72! contains (3 + 1) = 4 factors of 2.

- Factors of 5:
The prime factorization of 72! contains (1 + 0) = 1 factor of 5.

Number of Consecutive Zeroes:
To find the number of consecutive zeroes at the end of 72!, we need to determine the minimum number of times 2 and 5 appear as factors. Since there is only 1 factor of 5, we can conclude that there will be 1 zero at the end of 72!.

Therefore, the correct answer is option 'D' (1 zero).