Persistent Bugger

DESCRIPTION:

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.

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)

---

def persistence(n):
    count=0
    while n>=10:
        temp=1
        for i in str(n):
            temp*=int(i)
            print(temp)
        n=temp
        count+=1
    return count