Conjecture: Suppose that we have a two real functions, $f(n)\geq 0, g(n) \geq 0$ such that $2^{f(n)} \leq g(n) \leq 2^{a \cdot f(n)}$ with $a \geq 1$ and $f(n) $ is a strictly increasing function. This implies $g(n) = p(n) \cdot 2^{h(n)}$ when $n \in \mathbb{N}$ and $h(n) \neq 0$?
My thought is to prove it by using contradiction. Something along the lines of: Suppose that we have a two real functions, $f(n)\geq 0, g(n) \geq 0$ such that $2^{f(n)} \leq g(n) \leq 2^{a \cdot f(n)}$ with $a \geq 1$ and $f(n) $ is a strictly increasing function. Also, suppose that $g(n) \neq p(n) \cdot 2^{h(n)}$ when $n \in \mathbb{N}$ and $h(n) \neq 0$. Then do something to show that $g(n)$ is always divisible by 2, but not sure if this is a good method or how to apply the euclidean algorithm to something like this.
Since $2^{f(n)}\neq0$, for every $n\in\mathbb{N}$, let $$p(n)=\frac{g(n)}{2^{f(n)}},\ h(n)=f(n)$$ and consider that: $$g(n)=p(n)\cdot 2^{f(n)},\ \forall n\in\mathbb{N}$$
Note that $h(n)=f(n)\neq0$, since it is strictly increasing.