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?

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.
