Induction on $n$.Base case is obvious. Let's assume we have shown it for $1,2,...,n-1$.If we can delete some vertices to get a regular poly, then we are done. $\textbf{1)} n$ is even.The polygon is $\{P_{k}\}|1\leq k \leq n$ Then we can delete all the odd vertices and get a regular polygon. $\textbf{2)} n$ is odd.Let $v_2(n-1)=m \implies n=2^m.N+1$ where $N$ is odd.Now this process is to be followed. -Delete $P_{4k+3} \implies P_{4k+1}$ is left --Delete $P_{8k+5} \implies P_{8k+1}$ is left ---Delete $P_{16k+7} \implies P_{16k+1}$ is left $\vdots$ -Delete $P_{2^mk+(2^m+1)} \implies P_{2^mk+1}$ is left, which will finally give us a regular polygon.