Prove that if $n \in \mathbb{N}$ and $n \ge 2$, then $2^{n + 1} \le 3^n$. Prove that if $n \in \mathbb{N}$ and $n \ge 2$, then $2^{n + 1} \le 3^n$.
My method:
If $n = 2$, $2^{n + 1} \le 3^n$ then $2^3 \le 3^2$ is $8 \le 9$, which holds for $n = 2$. 
$2^{k + 1} \le 3^k$ then $2^{k + 2} \le 3^{k + 1}$. Then 
$2 \cdot 2 \cdot 2^k \lt 3 \cdot 3^k$
$4 \cdot 2^k \lt 3 \cdot 3^k$
Therefore, $4 \cdot 2^k \lt 3 \cdot 3^k$. Hence, $2^{n + 1} \le 3^n$ for $n \ge 2$.
Is there a problem with: $2^{k + 1} \le 3^k$ then $2^{k + 2} \le 3^{k + 1}$ or any other part of this proof?
 A: The proof is easy: 


*

*If $k=2$, we know that $2^{k+1} = 2^3 = 8 \leq 9 = 3^2 = 3^k$. Then, it is true. 

*We assume that it is also true for $k=n$, then we assume that $2^{n+1} \leq 3^{n}$

*We want to see that it is true if $k = n+1$:


$2^{k+1} = 2^{n+2} = 2^{n+1} \cdot 2$
We assumed before $2^{n+1} \leq 3^{n}$, then 
$2^{k+1} = 2^{n+1} \cdot 2 \leq 3^{n} \cdot 2 \leq 3^{n} \cdot 3 = 3^{n+1}=3^k \rightarrow 2^{k+1} \leq 3^k$
What we do is: 


*

*We prove that it is true for $k=2$

*We prove that if it is true for $k=n$, then it is also true for $k=n+1$


Therefore, it is true for $k=3$ because it is true for $k=2$. It is true for $k=4$ because it is true for $k=3$...
A: Formal proof by induction.

First, show that this is true for $n=2$:
$2^{2+1}<3^2$
Second, assume that this is true for $n$:
$2^{n+1}<3^n$
Third, prove that this is true for $n+1$:
$2^{n+2}=$
$2\cdot\color\red{2^{n+1}}<$
$2\cdot\color\red{3^n}<$
$3\cdot3^n=$
$3^{n+1}$

Please note that the assumption is used only in the part marked red.
