Problem: Let $A$ be a commutative ring with identity and $P$ be a prime ideal of the power series ring $A[[x]]$. Let $Q \subset A$ be the set of elements of the form $f(0)$, as $f(x)$ ranges over $P$. If $Q$ is generated by $n$ elements as an ideal of $A$, show that $P$ can be generated by $m$ elements as an ideal of $A[[x]]$, where $m = n$ if $x \notin P$ and $m = n+1$, if $x \in P$.
Attempt: I want to first look at the case when $x \notin P$. Suppose $Q$ is generated by elements $a_1,...,a_n \in A$: $Q = (a_1,...,a_n)$. I think $P = (a_1,...,a_n)$, generated as an ideal in $A[[x]]]$. The main problem I face is showing that $Q \subset P$, if that is even true. If I ignore that for now and focus on trying to prove the inclusion $P \subset (a_1,...,a_n)$, then I start with $f(x) = \sum_{m = 0}^\infty \alpha_m x^m \in P$. Since $f(0) = \alpha_0 \in Q$, there exist $r_1,...,r_n \in R$ such that $\alpha_0 = r_1a_1 + ... + r_na_n$. Then I guess $f(x) = a_1f_1(x) + ... + a_nf_n(x)$ where
$$f_1(x) = r_1 + \sum_{m = 1}^\infty \beta_{1,m}x^m$$ $$...$$ $$f_n(x) = r_n + \sum_{m = 1}^\infty \beta_{n,m}x^m$$
for some unknown $\beta_{i,j} \in R$ we need to solve. We then have
$$a_1f_1(x) + ... a_nf_n(x) = \underbrace{a_1r_1 + ... + a_nr_n}_{= \alpha_0} + \sum_{m = 1}^\infty (a_1\beta_{1,m} + ... + a_n\beta_{n,m})x^m$$
Now the next step is to find $\beta_{i,j}$ such that $a_1\beta_{1,m} + ... + a_n\beta_{n,m} = \alpha_m$ for all $m \geq 1$. This will work if $\alpha_m \in Q$, meaning that I need to show that every coefficient of a power series in $P$, not just the constant coefficients, are in $Q$. The strategy I have is to show that if the power series $\sum_{m = 0}^\infty \gamma_m x^m$ lies in $P$, then so does the power series $\sum_{m = 1}^\infty \gamma_m x^m = x \left(\sum_{m = 1}^\infty \gamma_m x^{m-1}\right)$, and then use the fact that $P$ is prime and $x \notin P$ to recursively show that $\alpha_m \in Q$. But I have been unable to show this. I wonder if I am in the right direction.