1. **Stating the problem:** Prove by induction that $16 \mid 9^{n+1} - 8n - 9$ for all integers $n \geq 0$. 2. **Base case:** For $n=0$, check if $16 \mid 9^{0+1} - 8\cdot0 - 9 = 9 - 0 - 9 = 0$. Since $16 \mid 0$, the base case holds. 3. **Inductive hypothesis:** Assume for some $k \geq 0$ that $16 \mid 9^{k+1} - 8k - 9$. This means there exists an integer $m$ such that: $$9^{k+1} - 8k - 9 = 16m$$ 4. **Inductive step:** We want to prove that $16 \mid 9^{k+2} - 8(k+1) - 9$. Start with: $$9^{k+2} - 8(k+1) - 9 = 9 \cdot 9^{k+1} - 8k - 8 - 9 = 9 \cdot 9^{k+1} - 8k - 17$$ Using the inductive hypothesis, substitute $9^{k+1} = 16m + 8k + 9$: $$9 \cdot (16m + 8k + 9) - 8k - 17 = 9 \cdot 16m + 9 \cdot 8k + 81 - 8k - 17$$ Simplify: $$144m + 72k + 81 - 8k - 17 = 144m + 64k + 64$$ Factor out 16: $$16 \cdot (9m + 4k + 4)$$ Since $9m + 4k + 4$ is an integer, the expression is divisible by 16. 5. **Conclusion:** By induction, $16 \mid 9^{n+1} - 8n - 9$ for all $n \geq 0$. **Note:** The troublesome $-1$ you encountered likely came from a miscalculation. The correct simplification shows the expression is divisible by 16 without remainder.