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?
 A: I'm not sure what you mean by “$\implies 1.3\ge1.3$”.

The proof should go like this.


*

*The statement is true for $k=1$; indeed, $(1.3)^1=1.3=1+0.3\cdot1$

*Suppose the statement holds for $n=k$, that is, $(1.3)^k\ge1+0.3k$; then 
$$
(1.3)^{k+1}=(1.3)^k\cdot 1.3\ge(1+0.3k)\cdot(1+0.3)=1+0.3k+0.3+0.9k\ge
1+0.3(k+1)
$$
so the statement holds also for $n=k+1$.
Actually, it's even easier to prove that, for $r>0$,
$$
(1+r)^n\ge1+rn
$$
and yours is the special case for $r=0.3$. (The hypothesis $r>0$ can actually be relaxed to $r>-1$.) This is known as Bernoulli inequality.
A: (EDIT: I realized after I posted that you asked for a proof by induction. My bad! The earlier answer posted by egreg does this, and is much simpler in my opinion.)
We want to show that $1.3^n \geq 1 + 0.3n$ for all $n > 0$. Let's look only at the left side for now. We can write
$$ 1.3^n = (1 + 0.3)^n.$$
Then by the binomial theorem we have that
$$ 1.3^n = (1 + 0.3)^n 
= \sum_{k=0}^n \binom{n}{k} 0.3^k
= 1 + 0.3 n + \sum_{k=2}^n \binom{n}{k} 0.3^k,$$
where we pulled the first two terms out of the summation so that we can directly compare it to the right side of the inequality. Thus, we have that $1.3^n \geq 1 + 0.3n$ provided $\sum_{k=2}^n \binom{n}{k} 0.3^k \geq 0$, which we know is true since all of the involved quantities are positive (or zero).
A: 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 $x\ge -1$. 
A: This is true in more general form:
$$ (1+x)^n \geq 1+ nx,\quad x\geq 0$$
The proof is just the binomial theorem:
$$(1+x)^n = 1 + nx + \frac{n(n-1)}2x^2+\cdots \geq\, 1 + nx$$ just let $x = 0.3$.
This is obvious for $x\geq 0$, and for $x\geq -1$ is known as Bernoulli's inequality, as noted in the answer by egreg.
A: I want to comment on the use of $\Rightarrow$. Usually it is used to denote implications, not equivalences. Also the arrow should point into the direction of which statement you want to prove. Further $\Rightarrow$ is rather surrounded by statements, than by equations. For example "$X$ solves the equation x^2=3" and not only $x^2=3$. It is because $\Rightarrow$ relates statements which are true or false and an equation without context is not always true or false it depends on the context.
You wrote $n=k \Rightarrow \cdots \Rightarrow 1.3 \ge 1.3$ those "statements" are not equivalent.
With the equations $n=1$, $n=k$ and $n=k+1$ you want to label the different cases. So it is a list of cases and I think it would be better to write $n=1:$, $n=k:$, $\dots$. For $n=1$ this would look like
$$ n=1: \quad  (1.3)^1 = 1.3 \ge 1.3 = 1 + (0.3)1. $$
To get from $n=k$ to $n=k+1$ you should take the equation for $n=k$ and try to get the equation for $n=k+1$ out of it. In this case multiplication is a good choice:
$$ n=k+1: \quad (1.3)^{k+1} = 1.3 (1.3)^k \ge 1.3 (1+ (0.3)k)= 1+ 0.3 + 1.3(0.3)k \ge 1 +(0.3)(k+1). $$
A: Just an other way
If you are interested by an other proof,
$$(1.3)^n-1=(1.3)^n-1^n=(1.3-1)(1+1.3+(1.3)^2+...+(1.3)^{n-1})=0.3(1+\underbrace{1.3}_{\geq 1}+\underbrace{(1.3)^2}_{\geq 1}+...+\underbrace{(1.3)^{n-1}}_{\geq 1})\geq 0.3(\underbrace{1+...+1}_{n\ times })=0.3n$$
what prove the claim.
