Prove that $(1 + x)^n \ge (1 + nx)$ for all natural numbers $n,$ where $x > -1.$

(N/A) Let $P(n)$ be the given statement,i.e.,$P(n): (1 + x)^n \ge (1 + nx)$ for $x > -1.$
We note that $P(n)$ is true when $n = 1,$ since $(1 + x) \ge (1 + x)$ for $x > -1.$
Assume that $P(k): (1 + x)^k \ge (1 + kx)$ for $x > -1$ is true. $(1)$
We want to prove that $P(k + 1)$ is true for $x > -1$ whenever $P(k)$ is true. $(2)$
Consider the identity $(1 + x)^{k+1} = (1 + x)^k(1 + x).$
Given that $x > -1,$ so $(1 + x) > 0.$
Therefore,by using $(1 + x)^k \ge (1 + kx),$ we have $(1 + x)^{k+1} \ge (1 + kx)(1 + x).$
i.e.,$(1 + x)^{k+1} \ge (1 + x + kx + kx^2).$ $(3)$
Here $k$ is a natural number and $x^2 \ge 0,$ so $kx^2 \ge 0.$ Therefore,$(1 + x + kx + kx^2) \ge (1 + x + kx).$
And so we obtain $(1 + x)^{k+1} \ge (1 + (1 + k)x).$
Thus,the statement in $(2)$ is established. Hence,by the principle of mathematical induction,$P(n)$ is true for all natural numbers.