Let $L|K$ be a finite field extension, $v$ a non-archimedean valuation and $w$ an extension to $L.$ If $\mathcal{O}_L$ is the integral closure of the valuation ring $\mathcal{O}_v$ of $v$ in $L,$ show that the localization $\mathcal{O}_{\mathfrak{B}}$ of $\mathcal{O}_L$ at the prime ideal $\mathfrak{B} = \{\alpha \in \mathcal{O}_L | w(\alpha) >0\}$ is the valuation ring $\mathcal{O}_w$ of $w.$

This is an exercise from Neukirch's book "Algebraic Number Theory", pg. 166 which I know how to solve. However, I feel that a more conceptual proof should be available but I seem unavailable to find one. Is there a "good" reason to see why the exercise is true? If $v$ is discrete, the exercise is obvious, and I would want some reason / solution which would be as convincing to me as I am about the veracity of the claim in that situation. I am sorry if this question is vague. I am OK with high-powered machinery to prove this theorem, in fact, I would be happy if it was used. I think (?) that we would be done if we could prove that the extension $\mathcal{O}_\mathfrak{B} \subset \mathcal{O}_w$ is integral. If we could prove directly that $\mathcal{O}_\mathfrak{B}$ is a valuation ring, we would also be done. But neither of these facts are obvious to me.

For sake of convenience, I include the solution I have at the moment:

It is easy to see that $\mathcal{O}_{\mathfrak{B}} \subset \mathcal{O}_w.$ Conversely, if $x \in \mathcal{O}_w$ let $$a_nx^n + a_{n-1}x^{n-1} + \cdots + a_1x + a_0 =0$$ be an algebraic equation for $x$ with coefficients in $K.$ Let $a_i$ be the coefficient farthest to the left with the lowest valuation. Let $b_j=a_j/a_i,$ and divide our equation by $a_i$ to get a new algebraic equation
$$b_nx^n + b_{n-1}x^{n-1} + \cdots + b_1x+b_0=0.$$ Note that $v(b_j) >0$ if $j >i$ and that $v(b_j) \geq 0$ for all $i.$ Divide the equation with $x^i$ to get
$$(b_nx^{n-i}+ \cdots + b_{n-i+1}x+1)+x^{-1}(b_{n-2}+ \cdots + b_0x^{-i+1})=0.$$
Let $w= (b_nx^{n-i}+ \cdots + b_{n-i+1}x+1)$ and $y= b_{n-2}+ \cdots + b_0x^{-i+1}.$ We then have that $x = -y/w.$ We now show that $y \in \mathcal{O}_L,$ and that $w \in \mathcal{O}_L, w \not \in \mathfrak{B},$ thus proving our inclusion. We note that $\mathcal{O}_L$ is the intersection in $L$ of all valuation rings containing $\mathcal{O}_v.$ Using this fact, one shows without too much trouble that our claim is true.