If $a$ $\in$ $\mathbb R$ $\ni$ $a > -1$, then ($\forall n$ $\in$ $\mathbb R$) ($(1+a)^n \ge 1 + na$)
My main concern is twofold: Firstly, I am concerned that constant $a$ in the proposition may force the proof into cases, even though some nmber crunching shows that this may not be true; lastly, I was wondering if the proof that I wrote is valid. Any hints or critiques will be enlightening.
Base Cases
Suppose $n = 1$. Then
\begin{align}(1+a)^n \ge 1+na \\ 1+a \ge 1+a\end{align}
Suppose $n = 2$. Then
\begin{align}(1+a)^n \ge 1+na \\ (1+a)^2 \ge 1+2a\end{align}
which is true, since $1+a > 0$.
Induction Hypothesis $-$Suppose $(1+a)^n \ge 1 + na$.
To show that $\left(1+a\right)^\left(n+1\right)$ $\ge 1+a(n+1)$, we begin as follows
$$\begin{align} \\ 1+a(n+1) & = (1 + an ) + a \tag{distribute and commute}\\[0.5em] & \le (1+a)^n + a \tag{ind. hyp.}\\[0.5em] & \le (1+a)^n(1+a) \\ & = \left(1+a\right)^\left(n+1\right) \end{align}$$
as desired.
No.
You want to show that $(1+a)^{n+1} \ge 1+(n+1)a $.
The induction hypothesis is $(1+a)^{n} \ge 1+na $.
Then
$\begin{array}\\ (1+a)^{n+1} &= (1+a)(1+a)^n \quad\text{(getting ready to use the induction hypothesis)}\\ &\ge (1+a)(1+na) \quad\text{(using the induction hypothesis and $1+a > 0$)}\\ &= 1+(n+1)a+na^2 \quad\text{(algebra)}\\ &\ge 1+(n+1)a \quad\text{(and we are done)}\\ \end{array} $