A matrix $M\in\mathbb{Z}^{d\times d}$ is called a dilation matrix if all its eigenvalues have magnitude greater than $1$. For a dilation matrix $M$, we define $$\Omega_M:=(M^{-T}\mathbb{Z}^d)\cap[0,1)^d,\quad \Gamma_M:=(M[0,1)^d)\cap \mathbb{Z}^d,$$ where $M^{-T}$ is the inverse transpose of $M$. It can be seen that both $\Omega_M$ and $\Gamma_M$ has $d_M:=|det(M)|$ elements, so we enumerate these two sets $$\Omega_M=\{\omega_1,\dots,\omega_{d_{M}}\},\quad \Gamma_M:=\{\gamma_1,\dots,\gamma_{d_M}\}$$ such that $\omega_1=\gamma_1=0$. Now I want to show the following identity: $$\sum_{j=1}^{d_M}e^{-2\pi i\gamma_j\cdot \omega}=d_M\delta_{0}(\omega),\quad\forall \omega\in\Omega_M,$$ where $\delta_0(x)=1$ if $x=0$ and $\delta_0(x)=0$ if $x\in\mathbb{R}^d\setminus\{0\}$.
The identity is rather straight forward in the univariate case $d=1$, as the summation in the above identity is simply a partial sum of the geometric series. But I have no idea how to prove the general case. Any help is greatly appreciated.