First we prove it's true for n=1

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

Now we assume it's true for n=k

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

We need to show that

(2)     1(1!)+2(2!)+3(3!)+...+(k+1)(k+1)! ≟ (k+2)!-1 

We add (k+1)(k+1)! to both sides of (1)

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

So the truth of (1) implies the truth of (2). So the induction is complete.

Edwin