参考解法：

#!/usr/bin/python
# -*- coding: UTF-8 -*-

s=0
def fact(n):
   if n==1:
      return 1
   return n*fact(n-1)

for n in  range(1,21):
   a=fact(n)
   s+=a
print(s)

参考解法：

#!/usr/bin/python
# -*- coding: UTF-8 -*-

sum1=0
for each in range(1,21):
    sum2=1
    while each>1:
        sum2=sum2*each
        each-=1
    sum1+=sum2
print(sum1)

参考方法：

#!/usr/bin/python3

n=int(input("Enter a number:"))
s=0
def cal(n):
    pro=1
    for i in range(1,n+1):
        pro=pro*i
    return pro
for i in range(1,n+1):
    if i!=n:
        print("%d! + "%i,end='')
    else:
        print("%d! = "%i,end='')
    s=s+cal(i)
print(s)

使用两个递归函数实现，其中jie()生成阶乘，sum()对阶乘求和

#!/usr/bin/python
# -*- coding: UTF-8 -*-

def jie(n):
    if n==1:
        return 1
    else:
        return n*jie(n-1)

def sum(n):
    if n==1:
        return jie(1)
    else:
        return jie(n)+sum(n-1)

#a=int(input("请输入一个正整数：\n"))
#print(sum(a))
print(sum(20))

Python3下的实现

#python 3
from functools import reduce
a = 0
for n in range(1, 21):
    a += reduce(lambda x, y: x * y, range(1, n + 1))
print(a)