Suppose, $ A_1A_2.....A_{2^k} $(clockwise) is the regular polygon. $ A_iA_{i+1}\equiv l_i $. Now note that reflections on $ A_1A_2,A_2A_3 $ is equivalent to the clockwise rotation centered at $ A_2 $ that maps $ A_2A_{2^k} $ to $ A_2A_4 $ and similar for the next pairs of reflections. So we consider the movement of $ A_{2^k} $ in this sequence of rotations centered at $ A_2,A_4,....,A_{2^k} $. Clearly, at the first rotation $ A_2A_{2^k} $ maps to $ A_2A_4 $ and the next rotation maps it to $ A_4A_6 $ and so on. So at the end it comes back to $ A_2A_{2^k} $ again. Now note that in this way we also get that if $ A_{2^k} $ maps to $ P $ finally. Then $ A_{2^k}P $ is equal to the perimeter of $ A_2A_4....A_{2^k} $. Also note that, the total rotation is by $ 0 $ degree. So if image of $ O $ is $ O' $ after all these rotations, then $ OO'=A_{2^k}P $ which is clearly less than the perimeter of the $ 2^k $-gon.

Suppose $z_2$ be reflection of $z_1$ in line $a_1a_2$. Then it’s easy to see, $z_2-a_1=e^{2i\theta}(\frac{|z_1|^2}{z_1}-\frac {|a_1|^2}{a_1})$. Where $\theta=arg(a_2-a_1)$. Now in general, if $z_{n+1}$ is reflection of $z_n$ in line $a_na_{n+1}$ then, $z_{n+1}=a_n+ e^{2i\theta(n)}(\frac{|z_n|^2}{z_n}-\frac {|a_n|^2}{a_n})$. Where $\theta(n)=arg(a_{n+1}-a_n)$.Now take $O=0$ and radius is one. So take $a_n=e^{i\alpha(n)}$ where $\alpha(n)=\frac {(n-1)\pi}{2^{k-1}}$. Assuming $z_n=x_n+iy_n$ for all $n$ we’ve $x_{n+1}-Cos(\alpha(n))=Cos(2\theta(n))(x_n-Cos(\alpha(n)))$ and $y_{n+1}-Cos(\alpha(n))=-Cos(2\theta(n))(y_n-Cos(\alpha(n)))$ . Now basically see, $\theta(n)=tan^{-1}(\frac {Sin(\alpha_{n+1})-Sin(\alpha(n))}{Cos(\alpha_{n+1})-Cos(\alpha(n))}$. And so, $\theta(n)=\frac {\pi+(2n-1)\pi/(2^{k-1})}{2}$.So $Cos(2\theta(n))=-Cos(\frac {(2n-1)\pi}{2^k})$. So now we’ve $x_{n+1}=Cos(\frac {n-1\pi}{2^{k-1}}) -Cos(\frac {(2n-1)\pi}{2^k})(x_n- Cos(\frac {n-1\pi}{2^{k-1}}) )$. Now taking, $f(n)= Cos(\frac {n-1\pi}{2^{k-1}})-Cos(\frac {n-1\pi}{2^{k-1}})$ and, $g(n)= Cos(\frac {(n-1)\pi}{2^{k-1}})\times Cos(\frac {(n-1)\pi}{2^{k-1}})$ we’ve $x_{n+1}=f(n)x_n+g(n)$ . Now, just easy induction gives, $x_{n+1}=f(n-1)f(n-2).....f(1)g(0)+f(n-1)...f(2)g(1)+...f(n-1)g(n-2)+g(n-1)$. Also we get similarly, $y_{n+1}= f'(n-1)f'(n-2)...f'(1)g(0)+f'(n-1)..f'(2)g(1)+...f'(n-1)g(n-2)-g(n-1)$ , where $f'(n)=- Cos(\frac {n-1\pi}{2^{k-1}})-Cos(\frac {n-1\pi}{2^{k-1}})$. So now we’ve $x^2_n+y^2_n$ for all $n$. Using this it’s not difficult to show $x^2_{2^k}+y^2_{2^k}\geq (2^{k+1}Sin(\frac {\pi}{2^{k-1}}))^2$. ( Hint to prove, induction on $k$).