SOLUTION: Prove that for all positive integers n, 1•1!+2•2!+3•3!+...+n•n!=(n+1)!-1

Prove:

1•1!+2•2!+3•3!+...+n•n! = (n+1)!-1

The proof is by induction.

It is true for n=1

1•1! = 1 and (1+1)!-1 = 2!-1 = 2-1 = 1

Assume k is such that the proposition is true for n ≦ k

We are assuming that 

(1)   1•1!+2•2!+3•3!+...+k•k! = (k+1)!-1

We must use (1) to show (2) 

(2)   1•1!+2•2!+3•3!+...+(k+1)•(k+1)! = ((k+1)+1)!-1 = (k+2)!-1

We start with

(1)   1•1!+2•2!+3•3!+...+k•k! = (k+1)!-1

We add (k+1)•(k+1)! to both sides

      1•1!+2•2!+3•3!+...+k•k!+(k+1)•(k+1)! = (k+1)!-1+(k+1)•(k+1)!

= (k+1)!+(k+1)•(k+1)!-1 =

Factor (k+1)! out of the first two terms:

(k+1)!•[1+(k+1)]-1 = (k+1)!•[1+k+1]-1 = (k+1)!•(k+2)-1 = (k+2)!-1

So we have shown (2)

We know that it is true for n=1, and for n=k=1 we know that it is true for
n=k=2.  That means we know it is true for n=k=3, and that means we know it
is true for n=k=4, and so on.

QED

Edwin