1. The problem: Prove by mathematical induction that for all integers $n \geq 1$, $$\sum_{k=1}^n \log(k) = \log(n!).$$ 2. Formula and rules: We use the principle of mathematical induction which involves two steps: - Base case: Verify the statement for $n=1$. - Inductive step: Assume the statement is true for $n=m$, then prove it for $n=m+1$. 3. Base case ($n=1$): $$\sum_{k=1}^1 \log(k) = \log(1) = 0,$$ and $$\log(1!) = \log(1) = 0,$$ so the base case holds. 4. Inductive hypothesis: Assume $$\sum_{k=1}^m \log(k) = \log(m!).$$ 5. Inductive step: Show that $$\sum_{k=1}^{m+1} \log(k) = \log((m+1)!).$$ Starting from the left side, $$\sum_{k=1}^{m+1} \log(k) = \left(\sum_{k=1}^m \log(k)\right) + \log(m+1).$$ Using the inductive hypothesis, $$= \log(m!) + \log(m+1) = \log(m! \times (m+1)) = \log((m+1)!).$$ 6. Conclusion: By induction, the formula holds for all integers $n \geq 1$. --- Additional practice questions: 1. Prove by induction that for all $n \geq 1$, $$\log(2) + \log(4) + \cdots + \log(2n) = \log(2^n n!).$$ 2. Prove by induction that for all $n \geq 1$, $$\sum_{k=1}^n k \log(k) \leq \frac{n(n+1)}{2} \log(n).$$