1. **State the problem:** We want to prove that if for some natural number $n$ and real number $x > -1$, the inequality $$ (1+x)^n \geq 1 + nx $$ holds, then it also holds for $n+1$, i.e., $$ (1+x)^{n+1} \geq 1 + (n+1)x. $$ 2. **Given:** Assume $$ (1+x)^n \geq 1 + nx $$ is true for some $n \in \mathbb{N}$ and $x > -1 \in \mathbb{R}$. 3. **Goal:** Show $$ (1+x)^{n+1} \geq 1 + (n+1)x $$ under the same conditions. 4. **Proof:** Start from the left-hand side of what we want to prove: $$ (1+x)^{n+1} = (1+x)^n \cdot (1+x). $$ Using the inductive assumption: $$ (1+x)^{n+1} \geq (1 + nx)(1 + x). $$ Expand the right side: $$ (1 + nx)(1 + x) = 1 + x + nx + nx^2 = 1 + (n+1)x + nx^2. $$ Since $x > -1$ and $n \geq 1$, we have $nx^2 \geq 0$ because $x^2 \geq 0$ for all real $x$. Therefore, $$ 1 + (n+1)x + nx^2 \geq 1 + (n+1)x. $$ So, $$ (1+x)^{n+1} \geq 1 + (n+1)x. $$ 5. **Conclusion:** By mathematical induction, the inequality $$ (1+x)^n \geq 1 + nx $$ holds for all $n \in \mathbb{N}$ and real $x > -1$. Note: The restriction $x > -1$ ensures both sides are defined and the inequality is valid.