If $f\left(x\right)=\cos x\cos 2x\cos 2^{2}x\cos 2^{3}x\mathrm{.......}\cos 2^{n-1}xandn>1,then{f}^{\text{'}}\left(\frac{{\displaystyle \pi }}{{\displaystyle 2}}\right)$

$f\left(x\right)=\cos x\cos 2x\cos 2^{2}x\cos 2^{3}x\mathrm{.......}\cos 2^{n-1}x$

$\Rightarrow f\left(x\right)=\frac{\sin 2^{n}x}{2^n\sin x}$

$\Rightarrow f^1\left(x\right)=\frac{2^n\cos 2^{n}x\sin x-\sin 2^{n}x\cos x}{2^n{\sin }^{2}x}$

$\Rightarrow f^1\left(\frac{{\displaystyle \pi }}{{\displaystyle 2}}\right)=\frac{2^n\cos 2^{n-1}\pi }{2^n}=\cos 2^{n-1}\pi ={(-1)}^{2^{n-1}}=1$