To prove the inequality \( n^n + 1 > (n+1)^n \) for all integers \( n \geq 1 \) using mathematical induction, we follow the usual steps of an induction proof: base case and inductive step.

### Step 1: **Base Case (n = 1)**

For \( n = 1 \), the inequality becomes:
\[
1^1 + 1 > (1+1)^1
\]
which simplifies to:
\[
1 + 1 > 2
\]
\[
2 > 2
\]
This is false. The base case fails at \( n = 1 \), so we need to adjust our base case. Let's try starting at \( n = 2 \).

For \( n = 2 \), the inequality becomes:
\[
2^2 + 1 > (2+1)^2
\]
which simplifies to:
\[
4 + 1 > 3^2
\]
\[
5 > 9
\]
This is also false. The inequality holds for higher values of \( n \), so starting from \( n = 3 \) may work.

Let's try \( n = 3 \):
\[
3^3 + 1 > (3+1)^3
\]
Simplifies to:
\[
27 + 1 > 4^3
\]
\[
28 > 64
\]
This is also false. Therefore, the given inequality \( n^n + 1 > (n+1)^n \) may not be correct for smaller values of \( n \). Would you like help revising the problem statement or further exploration on larger values? Let me know how you'd like to proceed.

### Follow-up Questions:
1. Would you like to explore larger values of \( n \) to see where the inequality holds?
2. Would you like to verify the original inequality or modify it slightly for better induction results?
3. Should we explore different proof methods besides induction for this inequality?
4. Would you like to check the specific cases for higher values of \( n \) numerically?
5. Should we rework the inequality to find a better base case or condition?

**Tip:** When using induction, always carefully check the base case as it can sometimes suggest the need to adjust the inequality or starting point.

Explore the steps to prove the inequality n^n + 1 > (n+1)^n using mathematical induction. This detailed solution includes the base case, inductive step, and key formulas. Suitable for students in Grades 11-12 studying advanced algebra and induction.

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