I have the following exercise:
$A$ is a Noetherian ring. Prove that every $f$ belonging to the ring of formal series over $A$ is nilpotent if and only if all its coefficients are nilpotent.
I did a implication from nilpotency to nilpotency of each coefficient - that does not require noetherianity. In the second way the implication should require noetherianity. I did it that way: the ring is Noetherian, so every ideal contains a power of its radical. So the zero ideal contains a power of its radical, so there exist $k$, a natural number, such that $k$ is maximal degree of nilpotency in $A$. Therefore, if every coefficient of $f$ is nilpotent, then $f^k$ is nilpotent. Am I right?