Explanation

(c)
We have $1.1! + 2.2! + 3.3! + \ldots.\ + n.\ n!$
$= \sum_{r = 1}^{n}{r.(r!) =}\sum_{r = 1}^{n}{\lbrack(r + 1)r! - r!\rbrack}$
$= \sum_{r = 1}^{n}{\lbrack(r + 1)! - r}!\rbrack$
$= (2! - 1!) + (3! - 2!) + ... + \lbrack(n + 1)! - n!\rbrack$
$= (n + 1)! - 1! = (n + 1)! - 1$