Let $T$ be an invertible ergodic transformation on a Lebesgue space $X$ and $O$ be the dyadic odometer on $(0,1)$. Is it true that $T\times O$ is ergodic if and only if $T^{2^n}$ is ergodic for every $n \geq 0$ ?

In fact, I am pretty sure this is true, because, unless I'm wrong, I have just realized that I have shown this result during my own research some time ago, but with an usual point of view (too long to explain here), and I would like to be sure I don't misinterpret.

I see how to prove: $$ T\times O \text{ ergodic} \quad \implies \quad T^{2^n} \text{ ergodic}. $$ For $n=0$, just say that the first factor of an ergodic product is ergodic. For $n=1$ it suffices to look at the transformation induced by $T \times O$ on $X \times [0, \frac12[$. It is clearly isomorphic to $T^2 \times O$. Then this transformation is ergodic when $T \times O$ is ergodic, and then the first factor $T^2$ is ergodic as well. For $n=2$, it suffices to look at the transformation induced by $T \times O$ on $X \times [0, \frac14[$, and so on.

But how to prove the converse ?