Bernoulli's inequality

Base case $n=1$. $(1+x)^1=1+x$, equality.

Inductive step. Suppose $(1+x)^n\ge1+nx$. Since $1+x\ge0$ (because $x\ge-1$), multiplying by $1+x$ preserves the inequality:
$$(1+x)^{n+1}\ge(1+nx)(1+x)=1+(n+1)x+nx^2\ge1+(n+1)x.$$
(The last inequality uses $nx^2\ge0$.)

Conclusion. The inequality holds for all $n\ge1$, with equality iff $n=1$ or $x=0$.

Remark. For $x\ge0$, this is the simplest nontrivial lower bound on $(1+x)^n$. It is used to compare exponential growth with linear growth.