We will prove this by verifying hypotheses (a), (b), and (c) of Lemma \ref{lemma-when-done}. \medskip\noindent Since $\mathcal{F}_n$ is locally free over $\mathcal{O}_U/f^n\mathcal{O}_U$ we see that we have short exact sequences $0 \to \mathcal{F}_n \to \mathcal{F}_{n + 1} \to \mathcal{F}_1 \to 0$ for all $n$. Thus condition (b) holds by Lemma \ref{lemma-topology-I-adic-f}. \medskip\noindent By induction on $n$ and the short exact sequences $0 \to A/f^n \to A/f^{n + 1} \to A/f \to 0$ we see that the associated primes of $A/f^nA$ agree with the associated primes of $A/fA$. Since the associated points of $\mathcal{F}_n$ correspond to the associated primes of $A/f^nA$ not equal to $\mathfrak m$ by assumption (3), we conclude that $M_n = H^0(U, \mathcal{F}_n)$ is a finite $A$-module by (5) and Local Cohomology, Proposition \ref{local-cohomology-proposition-kollar}. Thus hypothesis (c) holds. \medskip\noindent To finish the proof it suffices to show that there exists an $n > 1$ such that the image of $$ H^1(U, \mathcal{F}_n) \longrightarrow H^1(U, \mathcal{F}_1) $$ has finite length as an $A$-module. Namely, this will imply hypothesis (a) by Lemma \ref{lemma-ML-better}. The image is independent of $n$ for $n$ large enough by Lemma \ref{lemma-ML-local}. Let $\omega_A^\bullet$ be a normalized dualizing complex for $A$. By the local duality theorem and Matlis duality (Dualizing Complexes, Lemma \ref{dualizing-lemma-special-case-local-duality} and Proposition \ref{dualizing-proposition-matlis}) our claim is equivalent to: the image of $$ \text{Ext}^{-2}_A(M_1, \omega_A^\bullet) \to \text{Ext}^{-2}_A(M_n, \omega_A^\bullet) $$ has finite length for $n \gg 1$. The modules in question are finite $A$-modules supported at $V(f)$. Thus it suffices to show that this map is zero after localization at a prime $\mathfrak q$ containing $f$ and different from $\mathfrak m$. Let $\omega_{A_\mathfrak q}^\bullet$ be a normalized dualizing complex on $A_\mathfrak q$ and recall that $\omega_{A_\mathfrak q}^\bullet = (\omega_A^\bullet)_\mathfrak q[\dim(A/\mathfrak q)]$ by Dualizing Complexes, Lemma \ref{dualizing-lemma-dimension-function}. Using the local structure of $\mathcal{F}_n$ given in (4) we find that it suffices to show the vanishing of $$ \text{Ext}^{-2 + \dim(A/\mathfrak q)}_{A_\mathfrak q}( A_\mathfrak q/f, \omega_{A_\mathfrak q}^\bullet) \to \text{Ext}^{-2 + \dim(A/\mathfrak q)}_{A_\mathfrak q}( A_\mathfrak q/f^n, \omega_{A_\mathfrak q}^\bullet) $$ for $n$ large enough. If $\dim(A/\mathfrak q) > 3$, then this is immediate from Local Cohomology, Lemma \ref{local-cohomology-lemma-sitting-in-degrees}. For the other cases we will use the long exact sequence $$ \ldots \xrightarrow{f^n} H^{-1}(\omega_{A_\mathfrak q}^\bullet) \to \text{Ext}^{-1}_{A_\mathfrak q}( A_\mathfrak q/f^n, \omega_{A_\mathfrak q}^\bullet) \to H^0(\omega_{A_\mathfrak q}^\bullet) \xrightarrow{f^n} H^0(\omega_{A_\mathfrak q}^\bullet) \to \text{Ext}^0_{A_\mathfrak q}( A_\mathfrak q/f^n, \omega_{A_\mathfrak q}^\bullet) \to 0 $$ If $\dim(A/\mathfrak q) = 2$, then $H^0(\omega_{A_\mathfrak q}^\bullet) = 0$ because $\text{depth}(A_\mathfrak q) \geq 1$ as $f$ is a nonzerodivisor. Thus the long exact sequence shows the condition is that $$ f^{n - 1} : H^{-1}(\omega_{A_\mathfrak q}^\bullet)/f \to H^{-1}(\omega_{A_\mathfrak q}^\bullet)/f^n $$ is zero. Now $H^{-1}(\omega^\bullet_\mathfrak q)$ is a finite module supported in the primes $\mathfrak p \subset A_\mathfrak q$ such that $\text{depth}(A_\mathfrak p) + \dim((A/\mathfrak p)_\mathfrak q) \leq 1$. Since $\dim((A/\mathfrak p)_\mathfrak q) = \dim(A/\mathfrak p) - 2$ condition (6) tells us these primes are contained in $V(f)$. Thus the desired vanishing for $n$ large enough. Finally, if $\dim(A/\mathfrak q) = 1$, then condition (5) combined with the fact that $f$ is a nonzerodivisor insures that $A_\mathfrak q$ has depth at least $2$. Hence $H^0(\omega_{A_\mathfrak q}^\bullet) = H^{-1}(\omega_{A_\mathfrak q}^\bullet) = 0$ and the long exact sequence shows the claim is equivalent to the vanishing of $$ f^{n - 1} : H^{-2}(\omega_{A_\mathfrak q}^\bullet)/f \to H^{-2}(\omega_{A_\mathfrak q}^\bullet)/f^n $$ Now $H^{-2}(\omega^\bullet_\mathfrak q)$ is a finite module supported in the primes $\mathfrak p \subset A_\mathfrak q$ such that $\text{depth}(A_\mathfrak p) + \dim((A/\mathfrak p)_\mathfrak q) \leq 2$. By condition (6) all of these primes are contained in $V(f)$. Thus the desired vanishing for $n$ large enough.