**Proof.**
The equivalence $(1) \Leftrightarrow (2)$ follows from Proposition 3.2.6.11, the implication $(2) \Rightarrow (3)$ is a special case of Proposition 3.1.6.10, the implication $(3) \Rightarrow (4)$ follows from Proposition 3.2.6.9, the equivalence $(4) \Leftrightarrow (5)$ from Example 3.2.6.13, and the implication $(5) \Rightarrow (6)$ is immediate.

We will complete the proof by showing that $(6)$ implies $(1)$. Assume that $X$ is connected and let $x \in X$ be a vertex for which the homotopy groups $\pi _{n}(X,x)$ vanish for $n \geq 1$. Let $\sigma : \operatorname{\partial \Delta }^ n \rightarrow X$ be a morphism of simplicial sets for some $n \geq 0$; we wish to show that $\sigma $ is nullhomotopic. For $n \leq 1$, this follows immediately from our assumption that $X$ is connected. We may therefore assume without loss of generality that $n \geq 2$. Let $\sigma _0$ denote the restriction $\sigma |_{ \Lambda ^{n}_{n} }$. Since the horn $\Lambda ^{n}_{n}$ is weakly contractible, the morphism $\sigma _0$ is nullhomotopic (Variant 3.2.6.10). Using the connectedness of $X$ we see that $\sigma _0$ is homotopic to the constant morphism $\sigma '_0: \Lambda ^{n}_{n} \rightarrow \{ x\} \hookrightarrow X$. Applying the homotopy extension lifting property (Remark 3.1.5.3), we conclude that $\sigma $ is homotopic to a morphism $\sigma ': \operatorname{\partial \Delta }^ n \rightarrow X$ satisfying $\sigma '|_{\Lambda ^{n}_{n}} = \sigma '_0$. It will therefore suffice to show that $\sigma '$ is nullhomotopic. Note that $\sigma '$ factors as a composition

\[ \operatorname{\partial \Delta }^ n \twoheadrightarrow \operatorname{\partial \Delta }^ n / \Lambda ^{n}_{n} \simeq \Delta ^{n-1} / \operatorname{\partial \Delta }^{n-1} \xrightarrow {\tau } X, \]

where $\tau $ carries the base point of $\Delta ^{n-1} / \operatorname{\partial \Delta }^{n-1}$ to the point $x$. Since the homotopy group $\pi _{n-1}(X,x)$ vanishes, Example 3.2.6.13 guarantees that $\tau $ is nullhomotopic, so that $\sigma '$ is also nullhomotopic (Remark 3.2.6.8).
$\square$