Why is there a unqiue ideal $n$-simplex in $\mathbb H^n$ with largest volume for $n\geq 3$?

Of course, this is something one can easily prove by calculation, so let me say what type of answer I am looking for.

Recall that this is a crucial step in Gromov's proof of Mostow Rigidity, a (very rough) sketch of which is as follows. If we have two cocompact subgroups $\Gamma_1,\Gamma_2\subseteq\operatorname{Isom}(\mathbb H^n)$ which are isomorphic as abstract groups, then such an isomorphism must extend to a homeomorphism of their boundaries (as hyperbolic groups). In other words, we get a self-homeomorphism of $\partial\mathbb H^n$. But now since simplicial volume is a homotopy invariant, this homeomorphism must preserve the $(n+1)$-tuples of points giving simplices of maximal volume. Then one proves that any self-homeomorphism of $\partial\mathbb H^n$ with this property is in fact induced by an element of $\operatorname{Isom}(\mathbb H^n)$, so $\Gamma_1,\Gamma_2$ are conjugate (via this element) in $\operatorname{Isom}(\mathbb H^n)$.

Is there a high-brow proof of the fact in the title of this question? (that is, one which uses some rigidity results in Lie groups). A first step at answering this question would be to realize that the configuration space of $n+1$ points in $\partial\mathbb H^n$ modulo isometries is nontrivial iff $n\geq 3$. Now volume is some real-analytic function on this moduli space. Is there a nice explanation for why it miraculously has a unique global maximum (in fact, a unique local maximum!) (which thus implies Mostow rigidity)?

Why is there a unqiue ideal $n$-simplex in $\mathbb H^n$ with largest volume for $n\geq 3$?

For $n=3$, this is a standard calculation, and for larger dimensions is much harder (see Haagerup and Munkholm); I made the mistake of originally stating that this is easy for all $n$. Anyway, I'm wondering if there is a deeper reason for this fact (one which does not rely on calculation); let me say what type of answer I am looking for.

Recall that this is a crucial step in Gromov's proof of Mostow Rigidity, a (very rough) sketch of which is as follows. If we have two cocompact subgroups $\Gamma_1,\Gamma_2\subseteq\operatorname{Isom}(\mathbb H^n)$ which are isomorphic as abstract groups, then such an isomorphism must extend to a homeomorphism of their boundaries (as hyperbolic groups). In other words, we get a self-homeomorphism of $\partial\mathbb H^n$. But now since simplicial volume is a homotopy invariant, this homeomorphism must preserve the $(n+1)$-tuples of points giving simplices of maximal volume. Then one proves that any self-homeomorphism of $\partial\mathbb H^n$ with this property is in fact induced by an element of $\operatorname{Isom}(\mathbb H^n)$, so $\Gamma_1,\Gamma_2$ are conjugate (via this element) in $\operatorname{Isom}(\mathbb H^n)$.

Is there a high-brow proof of the fact in the title of this question? (that is, one which uses some rigidity results in Lie groups). A first step at answering this question would be to realize that the configuration space of $n+1$ points in $\partial\mathbb H^n$ modulo isometries is nontrivial iff $n\geq 3$. Now volume is some real-analytic function on this moduli space. Is there a nice explanation for why it miraculously has a unique global maximum (in fact, a unique local maximum!) (which thus implies Mostow rigidity)?