One fixes the counterclockwise orientation for polygons. Let $\omega_1\omega_2\cdots\omega_m$ ($m\geq 3$) be a polygon, where $\omega_i$'s are its vertices represented by complex numbers. It is standard to show that the area of the polygon is given by the imaginary part of the expression $$-\frac 1 2\sum_{i=1}^m\omega_i\overline{\omega_{i+1}},\qquad (1)$$ where $\omega_{m+1}=\omega_1.$

Since the vertices of a regular $n$-gon is obtained from the $n$-th roots of unity by suitable translation, rotation and scaling, one may assume that the vertices of $A_{i1}A_{i2}\cdots A_{in}$ are represented by the complex numbers $$A_{ij}=\alpha_i+\beta_i\zeta^{j-1},1\leq j\leq n,1\leq i\leq m,$$ where $n=2k~(k\geq 2),$ and $\zeta=e^{2\pi i/n}$ is an $n$-th root of unity. Note that $\zeta^k=-1.$

By (1), the area of $A_{11}A_{21}\cdots A_{m1}$ (resp. $A_{1~k+1}A_{2~k+1}\cdots A_{m~k+1}$) is represented as the imaginary part of $$-\frac 1 2\sum_{i=1}^m(\alpha_i+\beta_i)(\overline{\alpha_{i+1}}+\overline{\beta_{i+1}})$$
$$({\rm resp.} -\frac 1 2\sum_{i=1}^m(\alpha_i+\beta_i\zeta^k)(\overline{\alpha_{i+1}}+\overline{\beta_{i+1}\zeta^k}),$$ where $\alpha_{m+1}=\alpha_1$ and $\beta_{m+1}=\beta_1$. Their sum (before taking the imaginary part) equals
$$-\frac 1 2\sum_{i=1}^m[(\alpha_i+\beta_i)(\overline{\alpha_{i+1}}+\overline{\beta_{i+1}})+(\alpha_i-\beta_i)(\overline{\alpha_{i+1}}-\overline{\beta_{i+1}})]$$
$$=-\sum_{i=1}^m(\alpha_i\overline{\alpha_{i+1}}+\beta_i\overline{\beta_{i+1}}),\qquad (2)$$ where one uses that fact that $\zeta^k=-1.$

Now replacing $A_{i1}$ by $A_{ij}$ (resp. $A_{i~k+1}$ by $A_{i~j+k}$) for $1\leq i\leq m$ amounts to replacing $\beta_i$ by $\beta_i\zeta^{j-1}$ (resp. $\beta_i\zeta^k=-\beta_i$ by $\beta_i\zeta^{j+k-1}=-\beta_i\zeta^{j-1}$) for $2\leq j\leq n$. This has no effect on (2), since $$(\beta_i\zeta^{j-1})(\overline{\beta_{i+1}\zeta^{j-1}})=\beta_i\overline{\beta_{i+1}}, {\rm etc.}$$ The original assertion follows after taking the imaginary part of (2).