If $x\ge -1$ and $n\ge 1$, then

Proof

Let $P(n):$ "$(1+x{)}^n\ge 1+nx$".

$P(1):$
$(1+x{)}^1=1+x$, and
$1+1x=1+x$
so $P(1)$ is TRUE

Assume $P(k)$.
$P(k+1):$
$(1+x{)}^{k+1}=(1+x{)}^k(1+x)$
By $P(k)$, $(1+x{)}^k(1+x)\ge (1+kx)(1+x)$
so $(1+x{)}^{k+1}\ge 1+(k+1)x+k{x}^2\ge 1+(k+1)x$
so $P(k+1)$ is TRUE whenever $P(k)$.

By principle of mathematical induction, $P(n)$ for all $n\in \mathbb{N}$
QED.