Help with an inductive proof of $(1+a)^n\ge 1+na$ Let $a$ be a fixed real number such that $a > -1$.
Prove $(1+a)^n \geq 1+an$ for any $n\in\mathbb{N}$.
Base case.
$n = 1$.
$(1+a)^1 = 1+a(1)$ 
$\implies 1+a = 1+a$.
$1+a = 1+a(1)$.
so we can assume $P(n)$ is true when $n = 1$.
Want to show $P(n+1)$ holds.
$(1+a)^n = 1+an$
$(1+a)^n(1+a) = (1+an)(1+a)$
$(1+a)^{n+1} = 1+a+an+a^2n$
Notice $a^2n \geq 0$ since for $a > -1, a^2 > 0$.
$(1+a)^{n+1} \geq 1+a+an$
$(1+a)^{n+1} \geq 1+a(n+1)$.
so $P(n+1)$ holds.
Can someone scrutinize my base case? If I use $n=1$, how can I show that it's greater than or equal to, since the base case is just equal?
 A: If you know $A=B$, you know that $A \le B$ and $A \ge B$.
In fact, $(A = B) \iff (A \le B) \land (A \ge B)$
Therefore, your base case becomes:
$n = 1$.
$(1+a)^1 = 1+a(1) = 1+a$
$\implies(1+a)^1 \ge 1+a(1)$
$\therefore P(1)$ holds
For the rest of the proof:
$(1+a)^n \ge 1+an$
$\implies (1+a)^n(1+a)\ge (1+an)(1+a)$
$= 1+an + a + a^2n$
$= 1+a(n+1)+a^2n$
$a^2 \ge 0, \forall a \in \Bbb{R}$
$\therefore 1+a(n+1)+a^2n \ge 1+a(n+1), \forall a\in \Bbb{R}, \forall n \ge 0$
$\therefore P(1) \land (P(n) \implies P(n+1))$
$\therefore P(n), \forall n \in \Bbb{N}$
A: The case $n=1$ has already been addressed in the comments.
A couple of further issues in your work:


*

*It is not true that if $a>-1$, the $a^2>0$. In fact, $0>-1$. However, $a^2n\ge0$ holds regardless of any hypothesis stronger than, say, $n\in\Bbb N$ and $a\in \Bbb R$ (and "$a^2n\ge0$" is sufficient for the purpose of the proof). The point where you use $a>-1$ is just in the the implication $$(1+a)^n\ge 1+na\implies (1+a)^{n+1}\ge(1+a)(1+na)$$

*... which leads us to your second imprecision: the inductive step that you must prove is $$\forall n\ge1,\ \left[(1+a)^n\ge1+na\longrightarrow (1+a)^{n+1}\ge 1+(n+1)a\right]$$ while, formally, you proved the weaker predicate $$\forall n\ge1,\ \left[(1+a)^n=1+na\longrightarrow (1+a)^{n+1}\ge 1+(n+1)a\right]$$ The algebra is the same as you did, though, just with the "$\ge$" sign instead of "$=$".
