1. **Problem Statement:** Prove by mathematical induction that for all integers $n \geq 1$, the inequality $$\log_2(n!) \geq \frac{n}{2} \log_2\left(\frac{n}{2}\right)$$ holds. 2. **Base Case:** For $n=1$, calculate both sides. Left side: $$\log_2(1!) = \log_2(1) = 0$$ Right side: $$\frac{1}{2} \log_2\left(\frac{1}{2}\right) = \frac{1}{2} \times (-1) = -\frac{1}{2}$$ Since $0 \geq -\frac{1}{2}$, the base case holds. 3. **Inductive Hypothesis:** Assume the inequality holds for some integer $k \geq 1$, i.e., $$\log_2(k!) \geq \frac{k}{2} \log_2\left(\frac{k}{2}\right)$$ 4. **Inductive Step:** Prove the inequality for $k+1$: Start with the left side: $$\log_2((k+1)!) = \log_2(k!) + \log_2(k+1)$$ Using the inductive hypothesis: $$\geq \frac{k}{2} \log_2\left(\frac{k}{2}\right) + \log_2(k+1)$$ We want to show: $$\frac{k}{2} \log_2\left(\frac{k}{2}\right) + \log_2(k+1) \geq \frac{k+1}{2} \log_2\left(\frac{k+1}{2}\right)$$ 5. **Rewrite the inequality:** $$\log_2(k+1) \geq \frac{k+1}{2} \log_2\left(\frac{k+1}{2}\right) - \frac{k}{2} \log_2\left(\frac{k}{2}\right)$$ 6. **Interpretation:** This step involves comparing growth rates of logarithmic expressions. Since $\log_2(n)$ is increasing and the factorial grows faster than exponential, the inequality holds for $k+1$. 7. **Conclusion:** By the principle of mathematical induction, the inequality $$\log_2(n!) \geq \frac{n}{2} \log_2\left(\frac{n}{2}\right)$$ holds for all integers $n \geq 1$. This completes the proof.