Proving a Mathematical Induction Statement for the Product of Integers and Their Factorials

To prove the statement:

1 middot; 1! 2 middot; 2! 3 middot; 3! ... n middot; n! (n 1)! - 1

for all integers n geq 1 using the principle of mathematical induction, we will follow these steps:

Step 1: Base Case

First, let's check the base case when n 1:

1 middot; 1! 1

Now calculate the right-hand side:

(1 1)! - 1 2! - 1 2 - 1 1

Thus, the base case holds true:

1 middot; 1! (1 1)! - 1

Step 2: Inductive Step

Assume the statement is true for some integer k geq 1. That is, we assume:

1 middot; 1! 2 middot; 2! 3 middot; 3! ... k middot; k! (k 1)! - 1

We need to show that the statement is also true for k 1:

1 middot; 1! 2 middot; 2! 3 middot; 3! ... k middot; k! (k 1) middot; (k 1)! (k 2)! - 1

Using the inductive hypothesis, we can rewrite this as:

(k 1)! - 1 (k 1) middot; (k 1)! (k 1)! (1 k 1) - 1 (k 1)! (k 2) - 1

Now simplify (k 1)! (k 2):

(k 1)! (k 2) (k 2)! since (k 2)! (k 2)(k 1)!

Thus we have:

(k 1)! (k 2) - 1 (k 2)! - 1

Conclusion:

We have shown that if the statement holds for n k, it also holds for n k 1. Since the base case holds and the inductive step is verified by the principle of mathematical induction, the statement is true for all integers n geq 1.

Therefore, we conclude that:

1 middot; 1! 2 middot; 2! 3 middot; 3! ... n middot; n! (n 1)! - 1 for all n geq 1.

Example:

For n 1, 1 middot; 1! 1 2! - 1 1! - 1, which is true.

If 1 middot; 1! 2 middot; 2! ... k middot; k! (k 1)! - 1 is true for a certain n geq 1, then:

1 middot; 1! 2 middot; 2! ... k middot; k! (k 1) middot; (k 1)! (k 1)! - 1 (k 1) middot; (k 1)! (k 1)! (1 k 1) - 1 (k 1)! (k 2) - 1

and (k 1)! (k 2) (k 2)! since (k 2)! (k 2)(k 1)!.

Thus, (k 1)! (k 2) - 1 (k 2)! - 1, which proves the statement for n k 1.

Induction Base:

Let n 1. Then, 1 middot; 1! - 1 (1 1)! - 1 - 1 1! - 1.

Inductive Step:

Assume that for a certain n geq 1, it is indeed:

1 middot; 1! 2 middot; 2! 3 middot; 3! ... n middot; n! (n 1)! - 1

Then:

1 middot; 1! 2 middot; 2! ... n middot; n! (n 1) middot; (n 1)! (1 middot; 1! 2 middot; 2! ... n middot; n!) (n 1) middot; (n 1)! (n 1)! - 1 (n 1) middot; (n 1)! (n 1)! (1 (n 1)) - 1 (n 1)! (n 2) - 1 (n 2)! - 1

This completes the inductive step, thereby proving the statement for all n geq 1.