2020? But it's still 2015!

$\begin{aligned} f(0) & = \sqrt{5} - 1 \\ f(1) & = (\sqrt{5} + 1) (1^2) (\sqrt{5} - 1) & = (5^1 - 1)(1^2) & \color{#3D99F6}{= (5^{2^0}-1)(1!)^2} \\ f(2) & = (5^1 + 1) (2^2) (5^1 - 1)(1^2) & = (5^2 -1) (2!)^2 & \color{#3D99F6}{= (5^{2^1}-1)(2!)^2} \\ f(3) & = (5^2 + 1)(5^2 -1) (3!)^2 & = (5^4 - 1)(3!)^2 & \color{#3D99F6}{= (5^{2^2}-1)(3!)^2} \\ f(4) & = (5^4 + 1)(5^4 -1) (4!)^2 & = (5^8 - 1)(4!)^2 & \color{#3D99F6}{= (5^{2^3}-1)(4!)^2} \\ f(5) & = (5^8 + 1)(5^8 -1) (5!)^2 & = (5^{16} - 1)(5!)^2 & \color{#3D99F6}{= (5^{2^4}-1)(5!)^2} \\ \color{#3D99F6}{\Rightarrow f(n)} & \color{#3D99F6}{ = (5^{2^{n-1}} -1)(n!)^2 } \\ \Rightarrow f(2020) & = (5^{2^{2019}} -1)(2020!)^2 \end{aligned}$

$\Rightarrow p = 2^{2019} \quad \Rightarrow q = 2020 \quad \Rightarrow \dfrac{2^q}{p} = \dfrac {2^{2020}}{2^{2019}} = \boxed{2}$

The function is

$f(x+1) = (5^{2^{x-1}} + 1)(x+1)^{2}f(x)$

Move $f(x)$ to the left side

$\frac{f(x+1)}{f(x)} = (5^{2^{x-1}} + 1)(x+1)^{2}$

For $x=0$

$\frac{f(1)}{f(0)} = (5^{2^{-1}} + 1)(1)^{2}$

For $x=1$

$\frac{f(2)}{f(1)} = (5^{2^{0}} + 1)(2)^{2}$

For $x=2$

$\frac{f(3)}{f(2)} = (5^{2^{1}} + 1)(3)^{2}$

Do it until $x=2019$

$\frac{f(2020)}{f(2019)} = (5^{2^{2018}} + 1)(2020)^{2}$

Mutiply all the functions

$\frac{f(1)}{f(0)} \times \frac{f(2)}{f(1)} \times ... \times \frac{f(2020)}{f(2019)} = (5^{2^{-1}} + 1)(1)^{2} \times (5^{2^{0}} + 1)(2)^{2} \times ... \times (5^{2^{2018}} + 1)(2020)^{2}$

$\frac{f(1)}{f(0)} \times \frac{f(2)}{f(1)} \times ... \times \frac{f(2020)}{f(2019)} = (5^{2^{-1}} + 1)(5^{2^{0}} + 1)(5^{2^{1}} + 1)...(5^{2^{2018}} + 1) \times (1)^{2}(2)^{2}(3)^{2}...(2020)^{2}$

$\frac{f(1)}{f(0)} \times \frac{f(2)}{f(1)} \times ... \times \frac{f(2020)}{f(2019)} = (5^{2^{-1}} + 1)(5^{2^{0}} + 1)(5^{2^{1}} + 1)...(5^{2^{2018}} + 1) \times (1 \times 2 \times 3 \times ... \times 2020)^{2}$

$\frac{f(1)}{f(0)} \times \frac{f(2)}{f(1)} \times ... \times \frac{f(2020)}{f(2019)} = (5^{2^{-1}} + 1)(5^{2^{0}} + 1)(5^{2^{1}} + 1)...(5^{2^{2018}} + 1) \times (2020!)^{2}$

Then $f(1)$ until $f(2019)$ will be eliminated, $f(0)$ and $f(2020)$ will remain

$\frac{f(2020)}{f(0)} = (5^{2^{-1}} + 1)(5^{2^{0}} + 1)(5^{2^{1}} + 1)...(5^{2^{2018}} + 1) \times (2020!)^{2}$

Remember $f(0) = \sqrt{5} - 1 = 5^{\frac{1}{2}} - 1 = 5^{2^{-1}} - 1$

$\frac{f(2020)}{5^{2^{-1}} - 1} = (5^{2^{-1}} + 1)(5^{2^{0}} + 1)(5^{2^{1}} + 1)...(5^{2^{2018}} + 1) \times (2020!)^{2}$

Move $5^{2^{-1}} - 1$ to the right side

$f(2020) = (5^{2^{-1}} - 1)(5^{2^{-1}} + 1)(5^{2^{0}} + 1)(5^{2^{1}} + 1)...(5^{2^{2018}} + 1) \times (2020!)^{2}$

Because $(a+b)(a-b) = a^{2} - b^{2}$ so

$f(2020) = (5^{2^{-1}} - 1)(5^{2^{-1}} + 1)(5^{2^{0}} + 1)(5^{2^{1}} + 1)...(5^{2^{2018}} + 1) \times (2020!)^{2}$

$f(2020) = (5^{2^{0}} - 1)(5^{2^{0}} + 1)(5^{2^{1}} + 1)...(5^{2^{2018}} + 1) \times (2020!)^{2}$

$f(2020) = (5^{2^{1}} - 1)(5^{2^{1}} + 1)...(5^{2^{2018}} + 1) \times (2020!)^{2}$

until

$f(2020) = (5^{2^{2018}} - 1)(5^{2^{2018}} + 1) \times (2020!)^{2}$

$f(2020) = (5^{2^{2019}} - 1) \times (2020!)^{2}$

So, $p = 2^{2019}$ and $q = 2020$ , then

$\frac{2^{q}}{p} = \frac{2^{2020}}{2^{2019}} = \boxed{2}$

*cmiiw

$\large f(x+1) = (5^{2^{x-1}} + 1)(x+1)^{2}f(x)$

So, $f(x) = (5^{2^{x-2}} + 1)(x)^{2}f(x-1)$

$f(x) = (5^{2^{x-2}} + 1)(x)^{2}(5^{2^{x-3}} + 1)(x-1)^{2}f(x-2)$

$f(x) = (5^{2^{x-2}} + 1)(5^{2^{x-3}} + 1)(5^{2^{x-4}} + 1)...(5^{2^{-1}} + 1)(x!)^{2}f(0)$

$f(x) = (5^{2^{x-2}} + 1)(5^{2^{x-3}} + 1)(5^{2^{x-4}} + 1)...(5^{2^{-1}} + 1)(x!)^{2}(5^{2^{-1}} -1)$

$f(x) = (5^{2^{x-1}} - 1)(x!)^{2}$

With $x=2020$ , $q=2020$ and $p=2^{2019}$

$\frac{2^{q}}{p}=\frac{2^{2020}}{2^{2019}}=2$

It is easy to guess that the function by the recurence relation given.

By induction it is proven that $f(x) = (5^{2^{x-1}} - 1)(x!)^{2} \forall x \in N$ .

Thus by inserting $x = 2020$ we can easily ge $p = 2^{2019}$ and $q = 2020$ .

Therefore, $\frac{2^{q}}{p} = 2$ .