Persistent Bugger

Write a function, persistence, that takes in a positive parameter num and returns its multiplicative persistence, which is the number of times you must multiply the digits in num until you reach a single digit.

Example

For example (Input --> Output):

39 --> 3 (because 3*9 = 27, 2*7 = 14, 1*4 = 4 and 4 has only one digit)
999 --> 4 (because 9*9*9 = 729, 7*2*9 = 126, 1*2*6 = 12, and finally 1*2 = 2)
4 --> 0 (because 4 is already a one-digit number)

Solution

PYTHON

import math


def persistent_bugger(n):
    count = 0
    while len(str(n)) > 1:
        n = math.prod([int(i) for i in str(n)])
        count += 1
    return count


print(persistent_bugger(39))