= Solution
A <Dedekind domain> is a <Noetherian> <integrally closed domain> in which every nonzero <prime ideal> is maximal. The equivalent local characterization is that its localizations at nonzero <prime ideals> are <discrete valuation rings>. A field is harmless as the dimension-zero case.
Write $R$ for the domain, $F$ for its <fraction field> and $I$ for the nonzero <fractional ideal>. Its proposed inverse is an $R$-submodule of $F$. Choose $0\ne d\in R$ with $dI\subseteq R$, so $d\in I^{-1}$ and the inverse is nonzero. Choose also $0\ne a\in I$. Every $x\in I^{-1}$ satisfies $xa\in R$, hence $I^{-1}\subseteq a^{-1}R$. Writing $a=r/s$ with $r,s\in R$ gives $rI^{-1}\subseteq sR\subseteq R$. Thus $I^{-1}$ is a <fractional ideal>; the <Noetherian> hypothesis also makes it finitely generated.
For a nonzero maximal <prime ideal> $\mathfrak p$, the <discrete valuation ring> $R_{\mathfrak p}$ makes $I_{\mathfrak p}$ principal, say $I_{\mathfrak p}=a_{\mathfrak p}R_{\mathfrak p}$. We need to justify that inversion commutes with this <localization>. One inclusion is immediate. Conversely, if $xI_{\mathfrak p}\subseteq R_{\mathfrak p}$, finite generation of $I$ supplies a common $s\notin\mathfrak p$ such that $sxI\subseteq R$. Therefore $sx\in I^{-1}$, proving
$$
(I^{-1})_{\mathfrak p}=(I_{\mathfrak p})^{-1}=a_{\mathfrak p}^{-1}R_{\mathfrak p}.
$$
It follows that $(II^{-1})_{\mathfrak p}=R_{\mathfrak p}$ at every such <prime ideal>. Globally $II^{-1}\subseteq R$. If this were a proper <ideal>, it would be contained in a maximal <ideal> and would remain proper after localizing there, a contradiction. In the field case every nonzero <fractional ideal> is the whole field. Hence in all cases
$$
\boxed{II^{-1}=R.}
$$
This proves that every nonzero <fractional ideal> of a <Dedekind domain> is an <invertible fractional ideal>, rather than presupposing its inverse exists.
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