I've created a short program to calculate Additive Persistence (I hope I still remembers it correctly), I think you could just do the same to Multiplication Persistence.
#include <stdio.h>
#include <stdlib.h>
int AdditivePersistence(int number)
{
int sum = 0;
int i;
for (i = number; i > 0; i /= 10)
sum += i % 10;
return number == sum ? 0 : (1 + AdditivePersistence(sum));
}
int main(void)
{
int number = 2718;
printf("Additive Persistence of %d is %d.\n", number, AdditivePersistence(number));
return 0;
}