1. Stating the problem: Prove that if $$ (1 + x)^n \geq 1 + nx $$ for $$ x > -1 $$ and $$ n \in \mathbb{N} $$, then $$ (1 + x)^{n+1} \geq 1 + (n+1)x $$. 2. Assume the induction hypothesis: $$ (1 + x)^n \geq 1 + nx $$. 3. To prove: $$ (1 + x)^{n+1} \geq 1 + (n+1)x $$. 4. Begin by expanding the left-hand side: $$ (1 + x)^{n+1} = (1 + x)^n (1 + x) $$ 5. Using the induction hypothesis, substitute: $$ \geq (1 + nx)(1 + x) $$ 6. Multiply out the right-hand side: $$ = 1 + x + nx + nx^2 = 1 + (n+1)x + nx^2 $$ 7. Since $$ x > -1 $$ and $$ n \in \mathbb{N} $$, the term $$ nx^2 \geq 0 $$ because $$ n \geq 1 $$ and $$ x^2 \geq 0 $$ for all real $$ x $$. 8. Therefore, $$ 1 + (n+1)x + nx^2 \geq 1 + (n+1)x $$ 9. Combining this with step 6: $$ (1 + x)^{n+1} \geq 1 + (n+1)x + nx^2 \geq 1 + (n+1)x $$ 10. Thus, $$ (1 + x)^{n+1} \geq 1 + (n+1)x $$, completing the induction step. 11. Conclusion: The inequality holds for all $$ n \in \mathbb{N} $$ and real $$ x > -1 $$. This proves that $$ (1 + x)^n \geq 1 + nx $$ implies $$ (1 + x)^{n+1} \geq 1 + (n+1)x $$ under the given conditions.