\[\begin{array}{l} {\text{Prove the identity:}} \hfill \\ \sum\limits_{k = 1}^n {\left( {{{\sin }^{2m}}\left( {\frac{{\pi k}}{{2n}}} \right) + {{\cos }^{2m}}\left( {\frac{{\pi k}}{{2n}}} \right)} \right)} = 2n \cdot \frac{{\left( {2m - 1} \right)!!}}{{\left( {2m} \right)!!}}. \hfill \\ m,n \in \mathbb{N} \hfill \\ \end{array} \]