Fraction Tower

We note that the $4$ -level fraction tower can be written as $\dfrac{\;\;\;8\;\;\;}{\dfrac{\;4\;}{\dfrac{2}{1}}} = \dfrac{8}{4}\cdot{}\dfrac{2}{1}$ .

Therefore, the $2017$ -level fraction tower is as follows:

$\begin{aligned} Q & = \frac{2^{2016}}{2^{2015}} \cdot{} \frac{2^{2014}}{2^{2013}} \cdot{} \frac{2^{2012}}{2^{2011}} \cdot{} ... \frac{2^6}{2^5} \cdot{} \frac{2^4}{2^3} \cdot{} \frac{2^2}{2^1} \\ & = \prod_{n=1}^{1008} \frac{2^{2n}}{2^{2n-1}} \\ & = \prod_{n=1}^{1008} 2 = 2^{1008} \end{aligned}$

$\Rightarrow A = \boxed{1008}$

Okay guys, the way I solved it was probably not the most efficient way, but it is pretty intuitive.
What I like the most is that it is easy to see the pattern here.

So the general answer of this tower fraction for $2^{n}$ , were $n \in \mathbb{N}$ is $\displaystyle 2^{\lceil \frac{n}{2} \rceil}$ .

Obs: Sorry for my handwriting, when I wrote this I was trying to make sense to myself, just later that I thought of posting this solution (regardless the fact that there were already other excellent solutions).

Obs2: $\lceil n \rceil$ denotes the ceiling function.

Let $f(x)$ the function defined as the fraction tower with $x$ at the top. $f(a^{b})=a^{\frac{b}{2}}$ if $b$ is an even number, so $f(2^{2016})=2^{1008}$