Let $A$ be a unique factorization domain and $x\in A$ a prime element, show that any prime ideal $(0)\subsetneq \frak p \subseteq$ $(x)$ is minimal 
Let $A$ be a unique factorization domain and $x\in A$ a prime element,
show that any for prime ideal $(0)\subsetneq \frak p \subseteq$ $(x)$
it already holds that $\frak p =$ $(x)$.

I was working through this proof and I would love to see whether I understood things correctly and if not, hopefully get some help. I.e. this is supposed to be a proof verification post.
Proof: Let $\frak p$ be a prime ideal in $A$ s.t. $(0)\subsetneq \frak{p} \subseteq$ $(x)$. Assume $x \not\in \frak p$ and let $p_0\in \frak p$ be a fixed element.
$\frak p \subseteq$ $(x)$ implies $p_0\in (x)$. Thus, there exists an element $p_1\in A$ s.t. $$p_0 = p_1x.$$ Since $\frak p$ is a prime ideal, we know that $p_1$ is required to be an element of $\frak p$ (since we're assuming $x\not\in \frak p$).
Analogously for $p_1$ there exists some element $p_2$ s.t. $p_1 = p_2x$ and by the same argument as above, it holds that $p_2\in \frak p$.
Since $\frak p$ is multiplicatively closed it holds that $$p_0p_1 = p_1xp_2x = p_1p_2x^2 \in \frak p.$$
Inductively we observe that for all $n\in \mathbb{N}$, $x^n$ divides $p_0$.
Since $p_1$ is a non-unit (being an element of $\frak p$) it has a unique factorization into prime elements. Since any such factorization is finite, the prime element $x$ can appear at most finitely many times which leads to a contradiction, therefore $x\in \frak p$. $\Box$

Now there are two arguments where i'm really uncertain. Is this claim

Since $\frak p$ is multiplicatively closed it holds that $$p_0p_1 =
 p_1xp_2x = p_1p_2x^2 \in \frak p.$$

valid?
Secondly, this claim

$x^n$ divides $p_0$

basically violates the unique factorization, doesn't it? Just want to make sure i'm not missing the key observation in this proof.
Thanks for any help and feedback!
 A: Your solution to this problem is more or less correct, and is very nice. Well done! But I think you're getting needlessly sidetracked a bit at the end, so let me see if I can rephrase your work in a way which makes the answer to your questions clear.
Let $\mathfrak{p}$ be a prime ideal which strictly contains $0$ and is contained in $(x)$. We want to show that $\mathfrak{p} = (x)$, so towards a contradiction, suppose $x \notin \mathfrak{p}$, and fix a nonzero element $p_{0} \in \mathfrak{p}$. By unique factorization, there is a maximal $k \in \mathbb{N}$ such that $x^{k}$ divides $p_{0}$, i.e. $x^{k}$ divides $p_{0}$ and $x^{k+1}$ does not divide $p_{0}$. This $k$ is the exponent of $x$ in the unique factorization of $p_{0}$.
Claim: For each $i \geqslant 1$, there is an element $p_{i} \in A$ such that $p_{0} = p_{i} x^{i}$.
Note that if we prove this claim, then this contradicts the maximality of $k$, and so we're done.
Proof: We proceed by induction. For the case $i = 1$, we note that $p_{0} \in (x)$, and so $p_{0} = p_{1}x$ for some $p_{1} \in A$.
Now, let $i > 1$, and suppose we have constructed $p_{i-1} \in A$ such that $p_{0} = p_{i-1}x^{i-1}$. Since $\mathfrak{p}$ is prime, and $x \notin \mathfrak{p}$, it follows that $x^{n}$ is not in $\mathfrak{p}$ for any $n \in \mathbb{N}$. Hence, $x^{i-1}$ is not in $\mathfrak{p}$, and so we must have $p_{i-1} \in \mathfrak{p}$, whence $p_{i-1} = p_{i}x$ for some $p_{i} \in A$. This $p_{i}$ is the desired element in the induction statement, since
$$p_{0} = p_{i-1}x^{i-1} = (p_{i}x)x^{i-1} = p_{i}x^{i}.$$
To concretely answer your questions, it's certainly true that $p_{0}p_{1} \in \mathfrak{p}$, because $\mathfrak{p}$ is an ideal, and $p_{0}, p_{1}$ are in $\mathfrak{p}$. (In fact, this would be true if only one of $p_{0}, p_{1}$ were in $\mathfrak{p}$.) But this is moot, because as you can see, you don't actually use this claim in your proof.
As for your second question, writing that "$x^{n}$ divides $p_{0}$ violates unique factorization" is a bit vague. It's true that this is the case, but you want to make it clear that the problem is that $x^{n}$ divides $p_{0}$ for all $n \in \mathbb{N}$, and the property of unique factorization that this violates is the maximality statement we use in the proof above. Hopefully, phrasing the proof in this way makes it clear how to express this contradiction.
