For any \(n \in \mathbb{N}\) and \(x > -1\), Bernoulli's inequality asserts that:

\[(1 + x)^n \geq 1 + nx.\]

The proof can be done using different methods, and here we will use mathematical induction.

Base case: For \(n = 1\), the inequality becomes:

\[(1 + x)^1 \geq 1 + 1 \cdot x,\]

which is true since \( (1 + x)^1 = 1 + x \).

Inductive step: Assume the inequality holds for some \(n \in \mathbb{N}\), meaning:

\[(1 + x)^n \geq 1 + nx.\]

We now prove it holds for \(n + 1\). Multiplying both sides of the assumed inequality by \(1 + x\) (which is positive for \(x > -1\)) gives:

\[(1 + x) \cdot (1 + x)^n \geq (1 + x) \cdot (1 + nx).\]

Expanding both sides yields:

\[(1 + x)^{n+1} \geq 1 + nx + x + nx^2 = 1 + (n+1)x + nx^2.\]

Since \(nx^2 \geq 0\), we have:

\[1 + (n+1)x + nx^2 \geq 1 + (n+1)x.\]

Therefore, we conclude:

\[(1 + x)^{n+1} \geq 1 + (n+1)x.\]

By the principle of mathematical induction, Bernoulli's inequality holds for every \(n \in \mathbb{N}\).