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**rcs** im trying to answer or show the proof but unfortunately there things that are not clear to me

Problem:

Find the sum of the series 1 (1!) + 2 (2!) + 3 (3!) + ... + n (n!)

Define n! = n (n - 1) (n - 2) ( n - 3) ... (3)(2)(1)

Example 8! = 8(7)(6)(5)(4)(3)(2)(1)

My attempt:

=> 1 (1!) + 2(2!) + 3(3!) + 4 (4!) +...+ (n-1) (n-1)! + n(n!) Row 1

add 1 => 2 (1!) + 3(2!) + 4(3!) + 5 (4!) ... + n - 1+ 1(n-1)! + (n+1)(n!) Row 2

Subtract (Row 2 - Row 1)

2 (1!) + 3(2!) + 4(3!) + 5 (4!) ... + (n-1+ 1)(n-1)! + (n+1)(n!)

- __1 (1!) + 2(2!) + 3(3!) + 4 (4!) +...+ (n-1) (n-1)! + n(n!) __

1 (1!) + 1(2!) + 1 (3!) + 1 (4!) + ... 1 (n -1)! +n (n!) Row 3 ( answer)

Subtract the values of Row 2 and Row 3, then

**2 (1!)** +** 3(2!**) + **4(3!)** + **5 (4!)+** ... + (**n - 1+ 1)(n-1)!** + (n+1)(n!) Row 2

- [ 1 (1!) + ** 1(2!)** + **1 (3!)** + **1 (4!)** + **1(5!)+** ... **1 (n -1)!** +** n (n!) **

therefore

(n+1)(n!) - 1 (1!)

**(n+1)! - 1 this is my answer**

Please correct me if im wrong

Thanks