= Solution
A <Noetherian module> satisfies the <ascending chain condition> on its <submodules>; equivalently every <submodule> is a <finitely generated module>. If the latter condition holds, the union of an ascending chain has a finite generating set lying in one member, so the chain stabilizes. Conversely, a submodule with no finite generating set allows successive choices outside the submodule generated so far, producing a strictly ascending chain. A <Noetherian ring> is a ring that is <Noetherian> as a module over itself, equivalently one whose <ideals> are all finitely generated.
The <formal power series ring> consists of sequences written $\sum_{i\ge0}a_ix^i$, with coefficientwise addition and the <Cauchy product>:
$$
\left(\sum_{i\ge0}a_ix^i\right)\left(\sum_{j\ge0}b_jx^j\right)=\sum_{n\ge0}\left(\sum_{i=0}^na_ib_{n-i}\right)x^n.
$$
Every coefficient uses a finite sum; no analytic convergence is involved. Evaluation at zero gives $A[[x]]/(x)\cong A$. A <quotient ring> of a <Noetherian ring> is <Noetherian>, because its ascending chains of <ideals> lift to such chains upstairs. Thus one direction of <Noetherianity of a formal power series ring> is immediate.
For the other direction suppose $A$ is <Noetherian>, and let $I\subseteq A[[x]]$ be an <ideal>. Define the <coefficient ideals of a formal power series ideal>
$$
I_r=\{[x^r]g:g\in I\cap x^rA[[x]]\}\subseteq A.
$$
Each is an <ideal>, and multiplication by $x$ gives $I_r\subseteq I_{r+1}$. Choose $N$ with $I_N=I_{N+1}=\cdots$. For $0\le r\le N$, choose generators $a_{r1},\ldots,a_{r m_r}$ for $I_r$, and choose series $g_{rj}\in I\cap x^rA[[x]]$ whose $x^r$ coefficients are those generators. Empty generating sets are allowed for zero <ideals>.
These finitely many series generate $I$. To prove this without assuming that $I$ is closed, start with $g\in I$ and cancel its coefficients successively. If the residual $h_r$ lies in $I\cap x^rA[[x]]$, its $x^r$ coefficient lies in $I_r$. Put $s=\min(r,N)$; since $I_r=I_s$ when $r\ge N$, write that coefficient as $\sum_jc_{rj}a_{sj}$ and subtract $\sum_jc_{rj}x^{r-s}g_{sj}$. The new residual belongs to $I\cap x^{r+1}A[[x]]$.
For $s<N$, the generators $g_{sj}$ are used only at the single step $r=s$. For $s=N$, their accumulated multipliers are the genuine <formal power series> $\sum_{r\ge N}c_{rj}x^{r-N}$. Thus the construction yields the exact coefficientwise identity
$$
g=\sum_{s<N}\sum_jc_{sj}g_{sj}+\sum_j\left(\sum_{r\ge N}c_{rj}x^{r-N}\right)g_{Nj}.
$$
Every coefficient of the residual vanishes after finitely many steps. This is an ordinary finite generating expression in $A[[x]]$, not merely an approximation by elements of a smaller <ideal>. Therefore every <ideal> is finitely generated, proving
$$
\boxed{A\text{ is Noetherian}\iff A[[x]]\text{ is Noetherian}.}
$$
Finally, for the surjective <module endomorphism> $f$, the kernels $\ker f\subseteq\ker f^2\subseteq\cdots$ stabilize because $M$ is a <Noetherian module>. Take $N$ with $\ker f^N=\ker f^{N+1}$. If $f(m)=0$, surjectivity of $f^N$ gives $u$ with $f^N(u)=m$. Then $f^{N+1}(u)=0$, so $u\in\ker f^N$ and $m=0$. Hence $f$ is injective as well as surjective; its inverse is automatically $A$-linear. This proves \b[$f$ is an isomorphism], the statement that <Noetherian modules are Hopfian>.
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