Actually, there is an abstract-nonsense proof (although that doesn't make constructing the equivalence by hand any less of a good exercise).  It goes through the equivalence of presheaves on a category A with discrete fibrations over A, so $\widehat{A} \simeq \mathrm{DFib}(A) \subset (\mathrm{Cat}\downarrow A)$.  Note that DFib(A) is a full subcategory of $\mathrm{Cat}\downarrow A$: any map between discrete fibrations is a map of fibrations.

Furthermore, a composite of discrete fibrations is a discrete fibration, and conversely any map between discrete fibrations is *itself* a discrete fibration.  Thus, for any discrete fibration B → A, we have $\mathrm{DFib}(A)\downarrow B \simeq \mathrm{DFib}(B)$.  Thus, if $B = A\downarrow X$ is the category of elements of a presheaf $X\colon A^{op} \to \mathrm{Set}$, then we have

$(\widehat{A}\downarrow X) \simeq (\mathrm{DFib}(A)\downarrow B) \simeq \mathrm{DFib}(B) \simeq \widehat{B} = \widehat{A \downarrow X}$

See also [this question](http://mathoverflow.net/questions/23887/slices-of-presheaf-categories).