If $A^p$ is ergodic, does that make $A$ ergodic? And possible circular proof?

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I am reading Walters' Introduction to Ergodic Theory and am struggling to make sense of the proof of Theorem 1.10. I have an issue with understanding the overall structure of the proof, and with a component of the proof itself.

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Issue regarding the proof:

The sentence

So if $a_n \neq 0$... $\quad$ ... and so $f$ is constant a.e.

in my mind just doesn't follow. Doesn't this show that $A^p$ is ergodic, not $A$?

Issue regarding the structure:

At the beginning of the proof, an assumption is made and is later invoked to make a conclusion. How does saying

Suppose that whenever $\gamma A^n = \gamma$ for some $n \geq 1$ we have $\gamma \equiv 1$.

allow us to use that later on to prove that having the trivial character is the only $\gamma \in \hat{G}$ that satisfies $\gamma \circ A^n = \gamma$. It seems rather circular?

Appreciate any help.

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If $A^p$ is ergodic, then $A$ must be ergodic: the contrapositive of this statement says that if $A$ is not ergodic, then $A^p$ is not ergodic, which is easy to verify - every $A$-invariant function is also $A^p$ invariant.

Regarding the second issue: the assumption that "whenever $\gamma A^n=\gamma$ for some $n\geq 1$ we have $\gamma = 1$" is the second part of the "if and only if" assertion in the theorem, so it must be assumed to prove one direction of the "if and only if".