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:

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

 persistence(4) # returns 0, because 4 is already a one-digit number
```

Source: codewars | Difficulty: EASY
Tags: Fundamentals

Starter code: def persistence(n):
	

Solutions:
def persistence(n):
	n = str(n)
	count = 0
	while len(n) > 1:
		p = 1
		for i in n:
			p *= int(i)
		n = str(p)
		count += 1
	return count


def persistence(n):
	if n < 10:
		return 0
	mult = 1
	while n > 0:
		mult = n % 10 * mult
		n = n // 10
	return persistence(mult) + 1


def persistence(n):
	if str(n) == 1:
		return 0
	count = 0
	while len(str(n)) > 1:
		total = 1
		for i in str(n):
			total *= int(i)
		n = total
		count += 1
	return count


def multiply_digits(n):
	product = 1
	while n > 0:
		product *= n % 10
		n //= 10
	return product

def persistence(n):
	count = 0
	while n > 9:
		n = multiply_digits(n)
		count += 1
	return count