Proof.
We show that the functor \[ \textrm{Ar}^{(n-1)}(\mathrm{Cat}_{(\infty, {k})}) \rightarrow\mathrm{Cat}_{(\infty, {k})} \times_{\mathrm{Cat}_{(\infty, {n})}} \textrm{Ar}^{(n-1)}(\mathrm{Cat}_{(\infty, {n})})\] is surjective and fully faithful.
Surjectivity amounts to the following: For any \((\infty,k)\)-category \(\mathcal D\) equipped with an \((n-1)\)-faithful functor \(\mathcal C' \rightarrow\iota_n \mathcal D\) from an \((\infty,n)\)-category \(\mathcal C'\), there exits an \((\infty,k)\)-category \(\mathcal C\) with an \((n-1)\)-faithful functor \(\mathcal C\rightarrow\mathcal D\) which under \(\iota_n\) gets mapped to the original functor \(\mathcal C' \rightarrow\mathcal D\).
Define \(\mathcal C\coloneqq \mathrm{Fact}_{n-1}(\mathcal C' \rightarrow\iota_n \mathcal D\rightarrow\mathcal D)\) as the factorization with respect to the (\((n-1)\)-surjective, \((n-1)\)-faithful) factorization system in \(\mathrm{Cat}_{(\infty, {k})}\), and hence equipped with morphisms \(\mathcal C' \rightarrow\mathcal C\rightarrow\mathcal D\) where the former is \((n-1)\)-surjective and the latter is \((n-1)\)-faithful. To conclude, we show that the map \(\mathcal C' \simeq \iota_n \mathcal C' \rightarrow\iota_n \mathcal C\) is an equivalence, and hence that \(\iota_n(\mathcal C\rightarrow\mathcal D)\) is equivalent to \(\mathcal C' \rightarrow\iota_n \mathcal D\). Consider the following commutative diagram: 
The bottom horizontal and leftmost diagonal functor are surjective/faithful as indicate by the definition of \(\mathcal C\). The top-most diagonal functor is \((n-1)\)-faithful by assumption. The top horizontal functor is \((n-1)\)-faithful since \(\iota_n\) preserves faithfulness by observation 5.3.10. It then follows from lemma 5.3.5 that the functor \(\mathcal C'\rightarrow\iota_n \mathcal C\) is \((n-1)\)-faithful. Since \(\mathcal C' \rightarrow\mathcal C\) is \((n-1)\)-surjective and since \(\iota_n \colon \mathrm{Cat}_{(\infty, {k})} \rightarrow\mathrm{Cat}_{(\infty, {n})}\) preserves \((n-1)\)-surjective functors by lemma 5.3.11 it follows that \(\mathcal C' \simeq \iota_n \mathcal C' \rightarrow\iota_n \mathcal C\) is \((n-1)\)-surjective. Hence, \(\mathcal C' \rightarrow\iota_n \mathcal C\) is \((n-1)\)-faithful and \((n-1)\)-surjective and hence an equivalence.
We now prove that ([00BZ]) induces an equivalence on the hom-space between any pair of objects \(\{\mathcal C_1 \rightarrow\mathcal D_1\}, \{\mathcal C_2 \rightarrow\mathcal D_2\} \in\ \textrm{Ar}^{(n-1)}(\mathrm{Cat}_{(\infty, {k})})\), and hence that ([00BV]) is fully faithful. Unwinding the hom-spaces in the relevant arrow \(\infty\)-categories, this is equivalent to the statement that for any fixed \(\mathcal D_1 \rightarrow\mathcal D_2\) and any fixed dashed lift as shown in the first diagram in ([00C0]), the space of dashed lifts as shown in the commuting square in the second diagram in ([00C0]) is contractible. 
By lemma 5.3.14, \(\iota_n \mathcal C_1 \rightarrow\mathcal C_1\) is \((n-1)\)-surjective, and \(\mathcal C_2 \rightarrow\mathcal D_2\) is \((n-1)\)-faithful by assumption, hence the space of lift is contractible since \((n-1)\)-surjective/\((n-1)\)-faithful functors form a factorization system on \(\mathrm{Cat}_{(\infty, {k})}\). ◻