Use mathematical induction to prove the following statement:
For all $b\in\mathbb R$, and for all $n\in\mathbb N$, $$b>-1\implies (1+b)^n \geq 1+nb$$
When $n=1$, the inequality still holds $1+b \geq 1+b$.
For n+1$: $$(1+b)^{n+1} \geq 1+(n+1)b$$ Here I'm not sure the best way to simplify... $$(1+b)^n(1+b)\geq 1+bn+b$$
On
A few points:
The base case, in this case, is $n=0,$ so that's what you should verify, not $n=1.$
$$\color{green}{(1+b)^{n+1}} \equiv \underbrace{(1+b)^n(1+b) \geq (1+bn)(1+b)}_{\text{induction hypothesis}} \equiv 1+(n+1)b+\underbrace{b^2n}_{\geq 0} \color{green}{\geq 1+(n+1)b} .$$
Never say something like "let $n=n+1$". It makes no sense, algebraically. You can say that if the result is true for $n$, then it's true for $n+1$, or you can say "if the result is true for $n=k$, then it's true for $n=k+1.$"
For the induction step, if $(1+b)^n\geq 1+bn$, then $$ (1+b)^{n+1}-(1+b(n+1))\geq (1+b)(1+bn)-(1+b(n+1))=b^2n\geq 0. $$ Note that the first inequality above uses both the induction hypothesis and $b>-1$.