firstly let's find $f(x)$ let $g_n(x) = \frac{1}{g_{n-1}(x) + 1}$ and $g_0(x) = x$ so we have $g_1(x) = \frac{1}{1+x}, g_2(x) = \frac{1}{1+\frac{1}{1+x}}, ...$
notice that $g_n(x) + 1 = \frac{1}{g_{n+1}(x)}$
thus we have $(g_n(x))! = \Gamma(\frac{1}{g_{n+1}(x)})$ with some simplification we have $f(x) = \Gamma(\frac{1}{g_{2}(x)})\Gamma(\frac{1}{g_{3}(x)})...$
and with the below definition (Euler's) in mind :
$$
\Gamma(z) = \frac{1}{z} \prod_{n = 1}^\infty \frac{1}{1+\frac{z}{n}}(1+\frac{1}{n})^z
$$
we have :
$$
f_k(x) = (g_2(x)g_3(x)...g_k(x))\prod_{n = 1}^\infty \frac{1}{1+\frac{1}{ng_2(x)}}\frac{1}{1+\frac{1}{ng_3(x)}}...\frac{1}{1+\frac{1}{ng_k(x)}}(1+\frac{1}{n})^{g_1(x) + g_2(x) + ... + g_{k - 1}(x) + k - 1}
$$
also since we have $g_n(x) + 1 = \frac{1}{g_{n+1}(x)}$ :
$$
\frac{1}{1+\frac{1}{ng_k(x)}} = \frac{n}{n+1+g_{k-1}(x)}
$$
thus :
$$
f_k(x) = (g_2(x)g_3(x)...g_k(x))\prod_{n = 1}^\infty \frac{n}{n+1+g_{1}(x)}\frac{n}{n+1+g_{2}(x)}...\frac{n}{n+1+g_{k-1}(x)}(1+\frac{1}{n})^{g_1(x) + g_2(x) + ... + g_{k - 1}(x) + k - 1}
$$
note that we have :
$$
A_{2}(x) = g_2(x)g_3(x)...g_k(x) = \frac{g_2}{1+g_2}g_4g_5...g_k = (1 - \frac{1}{1+g_2})g_4g_5...g_k = (1-g_3)g_4g_5...g_k \Rightarrow A_{2} = A_{4} - A_{3}
$$
And generally $A_n = A_{n+2} - A_{n+1}$ moreover $A_{n} = A_{n-1} + A_{n-2}$
also note that $g_0(0) = 0$ and $g_1(0) = 1$, thus:
$$A_0 (0) = 0, A_1(0) = 1 \frac{1}{2} \frac{2}{3} \frac{3}{5} \frac{5}{8} \frac{8}{13}\frac{13}{21}...g_k(0)$$
did you notice the fibonacci sequence reaking?$A_{1,k} = \frac{1}{F_{k+1}}$
therefore $A_n(0) = \frac{F_n}{F_{k+1}}$.$$$$
from here finding the closed form shall be a practice for the reader as it is trivially derived.(Ask for further clarification.)
