Let P(n) be the given statement, that is,
P(n) : 1 × 1! + 2 × 2! + 3 × 3! + ... + n × n! = (n + 1)! – 1 for all natural numbers n.
Note that P (1) is true, since
P (1) : 1 × 1! = 1 = 2 – 1 = 2! – 1.
Assume that P(n) is true for some natural number k, i.e.,
P(k) : 1 × 1! + 2 × 2! + 3 × 3! + ... + k × k! = (k + 1)! – 1
To prove P (k + 1) is true, we have
P (k + 1) : 1 × 1! + 2 × 2! + 3 × 3! + ... + k × k! + (k + 1) × (k + 1)!
= (k + 1)! – 1 + (k + 1)! × (k + 1)
= (k + 1 + 1) (k + 1)! – 1
= (k + 2) (k + 1)! – 1 = ((k + 2)! – 1
Thus P (k + 1) is true, whenever P (k) is true. Therefore, by the Principle of Mathematical Induction, P (n) is true for all natural number n.
