How to end showing that a statement is true by using induction? I am practicing how to show true statements using induction. 
Since there are eventually different ways to show that a statement is true, I wanted to ask the following: When can you say that a true statement has been shown?
Example:

Prove that for any natural number n $\ge$ 2 we have $2^n > n+1$

Showing the base case is easy: 
$$ 2^2 > 2+1 $$
But now we have the induction step and I wanted to ask whether it is okay to stop where I stopped.
$$ 2^{n+1} > (n+1)+1  $$
$$ 2^n * 2 > n+2 $$
$$ 2^n > \frac{n+2}{2}$$
$$ 2^n > \frac{n+2}{2}$$
$$ 2^n > \frac{n}{2} +1 $$
Can I know say that it is true because $n \in \mathbb{N}$ and $n \ge 2$?
Or do I have to transform the equation in another way? I sometimes do not know when something has been shown and when the transformation is not enough to conclude that the statement is true.
 A: Your proof is upside-down: you should start with something which is true and end up with what you want to prove
So instead say something like 


*

*Given $2^n > n+1$  and $n \gt \frac n2$

*you have: $2^n > \frac n2 +1$

*and multiplying both sides by two: $2^{n+1}> n+2$

*and minor rearrangement: $2^{n+1} > (n+1)+1$

*and with the base case you can say by induction ...

A: The general method for induction is as follows:
1: Show that the $n$ case being true $\implies$ $n+1$ case is true.
2: Show that the "base case" is true, i.e, $n=1$ or $n=0$ or some other easy case. 
3: $n=1$ case being true implies the $n=2$ case, which implies the $n=3$ case, then $n=4,5,...$ and so on.
A: So letting $a = 2^n$ and $b = n + 1$, you have an assumption of the form $$a > b$$ and you want to prove $$2a > b + 1$$
For this you almost always want to use the transitivity of $>$, that is $x > y \text { and } y > z \text{ then } x > z$.  Since your goal is $2a > b+1$ that makes $x = 2a$ and $z = b+1$.  So we need to find a suitable $y$ to establish 2 assumptions:
$$2a > y$$
$$y > b + 1$$
Since we already have $2a > 2b$ (the inductive assumption) that suggests using $y = 2b$, which means proving $2b > b + 1$ is enough.  That is, just proving $2(n + 1) > (n + 1) + 1$ is enough to finish the inductive step of the problem.
