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?