I'm trying to understand an example for proof by mathematical induction from "Book of Proof (PDF)" (pg 158). I understand the basis step, but I'm not following the simplification in the inductive step, which I've copied below.
Suppose $\sum_{i=0}^{k}i.i!=(k+1)!-1$. Then:
$\sum_{i=0}^{k+1}i.i!=(\sum_{i=0}^{k}i.i!)+(k+1)(k+1)!$
= $((k+1)!-1)+(k+1)(k+1)!$
= $(k+1)!+(k+1)(k+1)!-1$
= $(1+(k+1))(k+1)!-1$
= $(k+2)(k+1)!-1$
= $(k+2)!-1$
= $((k+1)+1)!-1$
From lines 3 to 4 and from 5 to 6 it looks like $(k+1)!$ simplifies to $1$, but how?
Suppose $\sum_{i=0}^{k}i.i!=(k+1)!-1$. Then:
$\sum_{i=0}^{k+1}i.i!=(\sum_{i=0}^{k}i.i!)+(k+1)(k+1)!$
$((k+1)!-1)+(k+1)(k+1)!$
$(k+1)!+(k+1)(k+1)!-1$
Factor out $(k+1)!$ to get $(1+(k+1))(k+1)!-1$
$(k+2)(k+1)!-1$
Since (k+2)(k+1)! = (k+2), !$(k+2)!-1$
$((k+1)+1)!-1$