Prove by induction the particular inequality $\left(1.3\right)^n \ge 1 + \left(0.3\right)n$ for every $n \in \mathbb N$ $\left(1.3\right)^n \ge 1 + \left(0.3\right)n$ for every $n \in \mathbb N$
Not sure where I'm going wrong in my Algebra, but I assume it's because I'm adding an extra term.  Is the extra term unnecessary because it's not a summation I'm trying to prove? 
Starting the proof:
$n=1 \Rightarrow \left(1.3\right)^{1} \ge 1 + \left(0.3\right)(1) \Rightarrow 1.3 \ge 1.3$
$n=k \Rightarrow \left(1.3\right)^{k} \ge 1 + \left(0.3\right)(k) \Rightarrow 1.3 \ge 1.3$
$n=k+1 \Rightarrow \left(1.3\right)^{k}+\left(1.3\right)^{k+1} \ge 1 + \left(0.3\right)(k+1) \Rightarrow 1.3 \ge 1.3$
After trying some Algebra I end up with this:
$= 1 + \left(0.3\right)k+\left(1.3\right)\left(1.3\right)^{k} \ge 1 + \left(0.3\right)k + .3$
Any hints?

Let's prove Bernoulli's inequality by induction. Assume $x\ge -1$. 
Base step $n=1$: 
$$(1+x)^1 = 1+1\cdot x.$$ 
Now induction step.  Assume $(1+x)^n \ge 1+nx$. Then 
\begin{align*} (1+x)^{n+1} &= (1+x)^n (1+x) \\
& \ge (1+nx) (1+x)\\
&  = 1+(n+1)x +nx^2\\
&\ge 1+(n+1)x.
\end{align*} 
Note that the first inequality holds because of the induction assumption and since $x\ge -1$. 
