Final step of exercise 11.7 from Atiyah-Macdonald ($\dim A[x]=\dim A+1$) Ex. 11.7 from Atiyah-Macdonald is basically to prove $\dim A[x]=\dim A+1$ for $A$ noetherian.

From exercise 11.6, we get $\dim A[x]\geq\dim A+1$, so we are left to prove "$\leq$".

I've followed the hint and proved that $\text{ht}(p[x])=\text{ht}(p)$ for every $p\subset A$ prime, but I'm still not sure why this implies $\dim A[x]\leq\dim A+1$.

Here is what I've thought so far: 

if $P_1\subsetneq...\subsetneq P_k$ is a maximal chain in $A[x]$, then $P_k$ is maximal and $k=\dim A[x]$. Suppose that $m:=P_k\cap A$ is a maximal ideal of $A$. That way, $A/m$ is a field and $A[x]/m[x]\simeq A/m[x]$ is a P.I.D. Since $P_k$ is maximal, $\overline{P_k}\in A[x]/m[x]$ is maximal, so $\text{ht}(\overline{P_k})=1$, which means $P_k$ is minimal prime of $m[x]$. That way, $k=\text{ht}(P_k)=\text{ht}(m[x])+1=\text{ht}(m)+1\leq\dim A+1$.

The problem with what I did is that it depends on $P_k\cap A$ being maximal. Is this necessarily the case? I feel like this is crucial, otherwise the hint wouldn't work. How do I prove that?

If it isn't true, how can I conclude that $\dim A[x]\leq\dim A+1$?
We'll use the following fact which occurs in the hint to Exercise 11.6 p. 126 of the book:

($\star$) If $f:A\to A[x]$ is the natural embedding, then the fiber of
$$
f^*:\operatorname{Spec}(A[x])\to \operatorname{Spec}(A)
$$ 
over a prime ideal $\mathfrak p$ of $A$ is order isomorphic to the spectrum of
$$
k\otimes_AA[x]\simeq k[x],
$$ 
where $k$ is the residue field at $\mathfrak p$.

We denote the Krull dimension of the ring $A$ by $\dim A$, and the height of the prime ideal $\mathfrak p$ by $h(\mathfrak p)$.

Let's prove the inequality $\dim A[x]\le1+\dim A$. 

It suffices to prove 
$$
h(\mathfrak p)\le1+h(\mathfrak p^{\operatorname c})
$$ 
for all prime ideal $\mathfrak p$ of $A[x]$ by induction on $h(\mathfrak p^{\operatorname c})$.

If $h(\mathfrak p^{\operatorname c})=0$, the statement follows from ($\star$).

Assume $h(\mathfrak p^{\operatorname c})&gt;0$. It suffices to show

(a) $h(\mathfrak q)\le h(\mathfrak p^{\operatorname c})$ for all prime ideal $\mathfrak q$ of $A[x]$ strictly smaller than $\mathfrak p$.

Indeed, there is a chain 
$$
\mathfrak q_0\underset\ne\subset\cdots\underset\ne\subset\mathfrak q_{n-2}\underset\ne\subset\mathfrak q\underset\ne\subset\mathfrak p
$$ 
with $n=h(\mathfrak p)$, and we get $h(\mathfrak q)\ge n-1=h(\mathfrak p)-1$, that is, $h(\mathfrak p)\le1+h(\mathfrak q)$. 

Let $\mathfrak q$ be a prime ideal as in (a).

If $\mathfrak q^{\operatorname c}\underset\ne\subset\mathfrak p^{\operatorname c}$, we have $h(\mathfrak q^{\operatorname c})&lt;h(\mathfrak p^{\operatorname c})$, and the induction hypothesis implies $h(\mathfrak q)\le 1+h(\mathfrak q^{\operatorname c})\le h(\mathfrak p^{\operatorname c})$.

If $\mathfrak q^{\operatorname c}=\mathfrak p^{\operatorname c}$, we have $\mathfrak q=\mathfrak p^{\operatorname c}[x]$ by ($\star$). This implies $h(\mathfrak q)=h(\mathfrak p^{\operatorname c})$.
Consider the quotient map $q : A[x] \rightarrow A[x]/(x) \cong A$. Clearly $q(P_k) = P_k \cap A$, and if $q(P_k)$ not maximal then by Zorn's lemma $q(P_k) \subset Q$ where $Q$ is a maximal ideal of $A$. However, this implies $ P_k \subset q^{-1}(Q)$ where $q^{-1}(Q)$ is a prime ideal, a contradiction.