How to understand this simplification?

Question

I have a problem exercize in which the context isn't relevant to my confusion.What I want is to understand how the factorials in the solution are manipulated in order to simplify the equation.

$$ 1(1!) + 2(2!) + ... + n(n!) + (n+1)[(n+1)!] = (n+1)! - 1 + (n+1)[(n+1)!]$$

Then they simplify it the right side:

$$(n+1)! - 1 + (n+1)[(n+1)!]$$ They re-orgarnize the terms so nothing confusing here:

$$(n+1)! + (n+1)[(n+1)!] - 1$$

However this is confusing: $$(n+1)! + [1 + ( n + 1)] - 1$$ Now how did they go from $(n+1)[(n+1)!] -1 $ to $[1 + ( n + 1)] - 1$ ?

They then proceed: $$(n+1)![n+2] - 1$$ $$(n+2)[(n+1)!] - 1$$ $$(n+2) - 1$$ Now how did they go from $(n+2)[(n+1)!] - 1$ to $(n+2) - 1$ ?

Answer 1

Either you or the book are making lots of typos:

$1(1!) + 2(2!) + ... + n(n!) + (n+1)[(n+1)!] = (n+1)! - 1 + (n+1)[(n+1)!]$

$= (n+1)! + (n+1)[(n+1)!] - 1$. Now factor out the common term $(n+1)!$ to get:

$= (n+1)! [ 1 + (n+1)] - 1$. This is $(n+1)!$ TIMES $[ 1 + (n+1)]$ then $-1$; not $(n+1)!$ PLUS $[ 1 + (n+1)]$ then $-1$.

We then continue:

$= (n+1)! [n + 2] - 1$. And obviously $(n+1)!(n+2) = [1*2*3*...*(n+1)](n+2) = (n+2)!$ so we get:

$= (n+2)! - 1$. That is $(n+2)$FACTORIAL$ -1$ and not $(n+2)$DON'T DO NOTHIN'$ -1$.

And the final result is $=(n+2)! -1$.

......

IMO the hardest part to get is

$1*1! + 2*2! + ... +n*n! = (n+1)! -1$.

Is this the inductive step of a proof by induction?

Answer 2

Looks like we're proving the identity $\sum_{k=1}^n kk! = (n+1)! - 1$ by induction.

But there are some typographical errors. The algebra should go: $$ (n+1)!-1+(n+1)[(n+1)!]=(n+1)![1+(n+1)] - 1 = (n+1)!(n+2) - 1 = (n+2)!-1 $$