IID implies Ergodicity

Question

The environment space is given by $\Omega:=P^{\mathbb{Z}^{d}}$, where P contains the 2d-vectors serving as admissible transition probabilities. An Element $\omega \in \Omega$ is defined as $\omega:=(\omega(x))_{x\in \mathbb{Z}^{d}}$, and $\omega(x)=\{\omega(x,e):|e|=1, e\in \mathbb{Z}^d\}$. We endow $\Omega$ with the product-topology and let $\mathbb{P}$ be a probability measure on $(\Omega,B(\Omega))$. First Question If one assume, that the coordinate maps on the product space are iid under $\mathbb{P}$ then that means, that the $\omega(x)$ are iid under law $\mathbb{P}$? For the second question: We define now translation mapping $t_{y}:\Omega \rightarrow \Omega$ defined als $(t_{y}\omega)(x,e):=\omega(x+y,e)$. Then the author says, that under the iid-assumption one gets, that $(\Omega,B(\Omega),(t_{x})_{x\in\mathbb{Z}^d},\mathbb{P})$ is ergodic. Second Question: How to prove this? I know, that i need to show, that $t_{x}$ is measure preserving, and i have to show, that $I:=\{A \in B(\Omega): t_{x}^{-1}A=A\}$ is $\mathbb{P}$-trivial for all $x \in \mathbb{Z}^{d}$.