**Proof.**
Recall that there exists an isomorphism $\mathbf{P}^1_ k \to E$ such that the normal sheaf of $E$ in $X$ pulls back to $\mathcal{O}(-1)$. Then $H^0(E, \mathcal{O}_ E) = k$. We will denote $\mathcal{O}_ n(iE)$ the restriction of the invertible sheaf $\mathcal{O}_ X(iE)$ to $E_ n$ for all $n \geq 1$ and $i \in \mathbf{Z}$. Recall that $\mathcal{O}_ X(-nE)$ is the ideal sheaf of $E_ n$. Hence for $d \geq 0$ we obtain a short exact sequence

\[ 0 \to \mathcal{O}_ E(-(d + n)E) \to \mathcal{O}_{n + 1}(-dE) \to \mathcal{O}_ n(-dE) \to 0 \]

Since $\mathcal{O}_ E(-(d + n)E) = \mathcal{O}_{\mathbf{P}^1_ k}(d + n)$ the first cohomology group vanishes for all $d \geq 0$ and $n \geq 1$. We conclude that the transition maps of the system $H^0(E_ n, \mathcal{O}_ n(-dE))$ are surjective. For $d = 0$ we get an inverse system of surjections of rings such that the kernel of each transition map is a nilpotent ideal. Hence $A = \mathop{\mathrm{lim}}\nolimits H^0(E_ n, \mathcal{O}_ n)$ is a local ring with residue field $k$ and maximal ideal

\[ \mathop{\mathrm{lim}}\nolimits \mathop{\mathrm{Ker}}(H^0(E_ n, \mathcal{O}_ n) \to H^0(E, \mathcal{O}_ E)) = \mathop{\mathrm{lim}}\nolimits H^0(E_ n, \mathcal{O}_ n(-E)) \]

Pick $x, y$ in this kernel mapping to a $k$-basis of $H^0(E, \mathcal{O}_ E(-E)) = H^0(\mathbf{P}^1_ k, \mathcal{O}(1))$. Then $x^ d, x^{d - 1}y, \ldots , y^ d$ are elements of $\mathop{\mathrm{lim}}\nolimits H^0(E_ n, \mathcal{O}_ n(-dE))$ which map to a basis of $H^0(E, \mathcal{O}_ E(-dE)) = H^0(\mathbf{P}^1_ k, \mathcal{O}(d))$. In this way we see that $A$ is separated and complete with respect to the linear topology defined by the kernels

\[ I_ n = \mathop{\mathrm{Ker}}(A \longrightarrow H^0(E_ n, \mathcal{O}_ n)) \]

We have $x, y \in I_1$, $I_ d I_{d'} \subset I_{d + d'}$ and $I_ d/I_{d + 1}$ is a free $k$-module on $x^ d, x^{d - 1}y, \ldots , y^ d$. We will show that $I_ d = (x, y)^ d$. Namely, if $z_ e \in I_ e$ with $e \geq d$, then we can write

\[ z_ e = a_{e, 0} x^ d + a_{e, 1} x^{d - 1}y + \ldots + a_{e, d}y^ d + z_{e + 1} \]

where $a_{e, j} \in (x, y)^{e - d}$ and $z_{e + 1} \in I_{e + 1}$ by our description of $I_ d/I_{d + 1}$. Thus starting with some $z = z_ d \in I_ d$ we can do this inductively

\[ z = \sum \nolimits _{e \geq d} \sum \nolimits _ j a_{e, j} x^{d - j} y^ j \]

with some $a_{e, j} \in (x, y)^{e - d}$. Then $a_ j = \sum _{e \geq d} a_{e, j}$ exists (by completeness and the fact that $a_{e, j} \in I_{e - d}$) and we have $z = \sum a_{e, j} x^{d - j} y^ j$. Hence $I_ d = (x, y)^ d$. Thus $A$ is $(x, y)$-adically complete. Then $A$ is Noetherian by Algebra, Lemma 10.97.5. It is clear that the dimension is $2$ by the description of $(x, y)^ d/(x, y)^{d + 1}$ and Algebra, Proposition 10.60.9. Since the maximal ideal is generated by two elements it is regular.
$\square$

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