设S=1*2*3*.*n,求S不大于400000时的最大n

python设S=123*.*n,求S不大于400000时的最大n

#include<stdio.h>

int main()

{int n,s=1;

 for(n=1;s<=400000;)

   s*=++n;

 printf("最大的n是：%d",--n); 

 return 0;

}

def func(number):
    ret = 1
    for i in range(1, 1000000):
        ret = ret * i
        print(ret)
        if ret > number:
            return i - 1

print(func(5))