Proof by induction: $n$th Fibonacci number is at most $ 2^n$

I'm trying to find the proof by induction of the following claim: For all $n\in\mathbb N$, $\operatorname{fibonacci}(n) \le 2^n$

My Proof:

Base case: $n = 1$

$\operatorname{fibonacci}(1) \le 2^ 1$ is $1 \le 2$, true.

Base case holds

Inductive Hypothesis:

Assume true for $n = k$: $\operatorname{fibonacci}(k) \le 2^k$

Show True for $\operatorname{fibonacci}(k+1) \le 2^{k+1} $

$$\operatorname{fibonacci}(k) * k \le 2^k k$$ $$\operatorname{fibonacci}(k) \le 2^{k+1} $$ I get stuck here

Any help?

Assume that:

All Fibonacci numbers are positive. $(\star)$

Then observe that: \begin{align*} \text{fibonacci}(k + 1) &= \text{fibonacci}(k) + \text{fibonacci}(k - 1) \\ &< \text{fibonacci}(k) + \text{fibonacci}(k - 1) + \text{fibonacci}(k - 2) &\text{by }(\star)\\ &= \text{fibonacci}(k) + \text{fibonacci}(k) \\ &= 2 \cdot \text{fibonacci}(k) \\ &\leq 2 \cdot 2^{k} &\text{by the ind. hypothesis} \\ &= 2^{k + 1} \end{align*} as desired. $~~\blacksquare$

It is clear in the base case that $F_1\le 2^1$ and $F_2\le 2^2$. Then in the inductive step we see that \begin{align} F_n &= F_{n-1}+F_{n-2}\\ &\le 2^{n-1} + 2^{n-2}\\ &= 2^{n-2}(2 + 1)\\ &\le 2^{n-2}(4)\\ &=2^n. \end{align}