Can you deduce this?

KEEP CALM :-)

Let's generalise $\mathfrak{P}=\displaystyle\prod_{r=0}^{n}\left(1+\sec (2^{r}x)\right)$ . $\begin{aligned}\mathfrak{P}=&\displaystyle\prod_{r=0}^{n}\left(\dfrac{\cos (2^{r}x)+1}{\cos (2^{r}x) }\right) \\&=2^{n+1}\displaystyle\prod_{r=0}^{n}\left(\dfrac{\cos^2 (2^{r-1}x)}{\cos (2^{r}x) }\right) \\&=\left(\dfrac{2^{n+1}\cos(x/2)}{\cos(2^n x)}\right)\displaystyle\prod_{r=0}^{n}\left(\cos (2^{r-1}x)\right)\\&= \left(\dfrac{2^{n+1}\cos(x/2)}{\cos(2^n x)}\right)\left(\dfrac{\sin(2^{n+1}\frac x2)}{2^{n+1}\sin\frac x2}\right)\\&=\tan\left(2^{n}x\right)\cot\dfrac{x}2\end{aligned}$ Coming back to integration, substituting $\mathfrak P$ and using $1-\cos x=2\sin^2(x/2)$ and $1+\cos x=2\cos^2 (x/2)$ so that $\cot(x/2)$ term in $\mathfrak P$ gets canceled and hence integration is:

$\displaystyle\int\tan \left(2^nx\right)\mathrm{d}x$ $=-\dfrac{1}{2^n}\ln\left(\cos(2^nx)\right)+\underbrace{C}_0$

$g(x)=f_{(4)}(x)=-\dfrac{1}{16}\ln\left(\cos(16x)\right)$

$g(x)=1\implies x=\dfrac{1}{16}\cos^{-1}e^{-16}$

$\therefore\large |k|=16$

$\huge\therefore (16-9)!=7!=\boxed{5040}$