1. **Problem:** Simplify the expression $$\frac{n!}{(n-1)!}$$ and identify the correct answer from the options. 2. **Recall the factorial definition:** $$n! = n \times (n-1) \times (n-2) \times \cdots \times 1$$ 3. **Rewrite the numerator using factorial properties:** $$n! = n \times (n-1)!$$ 4. **Substitute into the expression:** $$\frac{n!}{(n-1)!} = \frac{n \times (n-1)!}{(n-1)!}$$ 5. **Cancel the common factor $(n-1)!$:** $$\frac{n \times \cancel{(n-1)!}}{\cancel{(n-1)!}} = n$$ 6. **Final answer:** $$\frac{n!}{(n-1)!} = n$$ --- 7. **Problem:** Understand the principle of mathematical induction and identify which set it applies to. 8. **Mathematical induction is a proof technique used to prove statements about natural numbers.** 9. **The process involves:** - Proving the base case (usually $P(1)$ or $P(0)$) is true. - Assuming $P(k)$ is true for some arbitrary natural number $k$. - Proving $P(k+1)$ is true based on the assumption. 10. **Therefore, the correct answer is:** Mathematical induction is used to prove statements about **natural numbers**. --- 11. **Summary of multiple-choice answers:** - For $$\frac{n!}{(n-1)!}$$ the answer is **B) n**. - Mathematical induction applies to **C) Natural numbers**. --- 12. **Regarding the induction steps mentioned:** - A) $P(k)$ is the induction hypothesis. - B) $P(k+1)$ is the step to prove. - C) $P(1)$ or $P(0)$ is the base case. - D) $P(n)$ for all $n$ is the conclusion after induction. These are the standard components of a mathematical induction proof.