**Proof.**
We first show that $(1)$ implies $(2)$. For every groupoid $\mathcal{G}$, Proposition 1.3.5.2 guarantees that the simplicial set $\operatorname{N}_{\bullet }( \mathcal{G} )$ is a Kan complex. To show that $\operatorname{N}_{\bullet }( \mathcal{G} )$ is a $1$-groupoid, we must prove that if $\sigma , \tau : \Delta ^{m} \rightarrow \operatorname{N}_{\bullet }( \mathcal{G} )$ are $m$-simplices for $m \geq 2$ which have the same restriction to some horn $\Lambda ^{m}_{i} \subset \Delta ^ m$, then $\sigma = \tau $. For $m > 2$, this is immediate (since $\Lambda ^{m}_{i}$ contains the $1$-skeleton of $\Delta ^{m}$). In the case $m=2$, we can identify $m$-simplices of $\operatorname{N}_{\bullet }(\mathcal{G} )$ with commutative diagrams

\[ \xymatrix@R =50pt@C=50pt{ & Y \ar [dr]^{g} & \\ X \ar [ur]^{f} \ar [rr]^{h} & & Z } \]

in the groupoid $\operatorname{\mathcal{G}}$. The desired result then follows from the observation that any two of the morphisms $f$, $g$, and $h$ determine the third.

The implication $(3) \Rightarrow (1)$ is immediate. We will complete the proof by showing that $(2)$ implies $(3)$. Assume that $X$ is a $1$-groupoid and let $\mathcal{G} = \pi _{\leq 1}(X)$ be its fundamental groupoid. We wish to show that the tautological map $u: X \rightarrow \operatorname{N}_{\bullet }( \mathcal{G} )$ is an isomorphism: that is, it is bijective on $m$-simplices for $m \geq 0$. The proof proceeds by induction on $m$. The case $m = 0$ is immediate from the definitions. For $m \geq 2$, we have a commutative diagram

\[ \xymatrix@R =50pt@C=50pt{ \operatorname{Hom}_{\operatorname{Set_{\Delta }}}( \Delta ^{m}, X ) \ar [r]^-{u \circ } \ar [d] & \operatorname{Hom}_{\operatorname{Set_{\Delta }}}( \Delta ^ m, \operatorname{N}_{\bullet }(\mathcal{G}) ) \ar [d] \\ \operatorname{Hom}_{\operatorname{Set_{\Delta }}}( \Lambda ^{m}_{0}, X) \ar [r]^-{u \circ } & \operatorname{Hom}_{\operatorname{Set_{\Delta }}}( \Lambda ^{m}_{0}, \operatorname{N}_{\bullet }( \mathcal{G} ) ) } \]

where the vertical maps are bijective (since $X$ and $\operatorname{N}_{\bullet }(\mathcal{G} )$ are $1$-groupoids) and the bottom horizontal map is bijective (by virtue of our inductive hypothesis); it follows that the upper horizontal map is bijective as well. It will therefore suffice to treat the case $m = 1$. Let $e, e': x \rightarrow y$ be edges of the simplicial set $X$ having the same source and target; we wish to show that if the homotopy classes $[e]$ and $[e']$ coincide (as morphisms in the category $\mathcal{G} = \pi _{\leq 1}(X)$), then $e = e'$. Let $\sigma $ be a $2$-simplex of $X$ which is a homotopy from $e$ to $e'$: that is, a $2$-simplex whose boundary is depicted in the diagram

\[ \xymatrix@R =50pt@C=50pt{ & y \ar [dr]^{\operatorname{id}_ y} & \\ x \ar [ur]^{e} \ar [rr]^{e'} & & y } \]

(see Definition 1.4.3.1). Let $\tau $ be the right-degenerate $2$-simplex $s^{1}_{1}(e)$. Then $\sigma $ and $\tau $ have the same restriction to the horn $\Lambda ^{2}_{1} \subset \Delta ^2$. Invoking our assumption that $X$ is a $1$-groupoid, we conclude that $\sigma = \tau $. In particular, we have $e' = d^2_{1}(\sigma ) = d^{2}_{1}( \tau ) = e$.
$\square$