Bernoulli's inequality

Base case. For $n=0$ both sides equal $1$.

Induction step. Assume $(1+x)^n\ge1+nx$. Since $1+x\ge0$, multiplying by $1+x$ keeps the inequality:
$$(1+x)^{n+1}\ge(1+nx)(1+x)=1+(n+1)x+nx^2\ge1+(n+1)x.$$

So the statement holds for $n+1$, and by induction for all $n\ge0$.