Help me solve this mathematical induction ( 1 * 1! + 2 * 2! + ... + (n - 1) * (n - 1)! = n! - 1 where n > 0)

Question

Recently I had a Discrete maths exam and came across the following induction. I improvised in the last step just for the points, but i cant seem to find the actual solution.

In case it is not clear the exercise is: $$1\cdot1!+2\cdot2!+\dots+(n-1)(n-1)!=n!-1\text{ where }n>0$$

Answer 1

If it's true for $n-1$ namely $1*1! + \dots +(n-2)(n-2)! = (n-1)! - 1$ than for $n$ we have $1*1! + \dots + (n-2)(n-2)! + (n-1)(n-1)! = (n-1)! - 1 + (n-1)(n-1)! = (n-1)! - 1 + n! - (n-1)! = n! - 1$

Q.E.D

Answer 2

Using induction

$$P(n):=\sum_{i=1}^{n-1}i\cdot i!=n!-1\\ P(n+1) = \sum_{i=1}^{n}i\cdot i!=(n+1)!-1 \\ P(n+1) := \sum_{i=1}^{n-1}i\cdot i! + n\cdot n!=n! - 1 + n\cdot n!=(n+1)!-1$$

Using telescopic sum

$$T(n) = n!-(n-1)! \implies \sum_{i=1}^nT(n) = n!-1$$

Answer 3

Step: $n+1:$

Consider the RHS for $n+1:$

$(n+1)!-1 =(n+1)n! -1=$

$ (n!-1) + (n)n!=$

(using the hypothesis)

$(1×1!+..(n-1)(n-1)!)+ (n)n! .$