I am not sure what you mean by a direct proof, but here is a reasonably straightforward argument. Consider your equi-angular simplex $S$, and consider the link $L(v)$ of a vertex $v$ (the intersection of a small sphere centered at $v$ with your simplex, scaled so that the sphere has radius $1.$ The dihedral angles of $L(v)$ are equal to the corresponding dihedral angles of $S,$ so the links of all vertices are equiangular. We can show that an equiangular spherical simplex is regular by polar duality (as you had suspected): The dual simplex $T^\ast$ of a spherical simplex $T$ is one whose vertices are the face normals of $T.$ This polar duality interchanges distances with exterior dihedral angles, and transforms the Gram matrix in a very simple way, described in this paper by Kokkendorff (2005). The regular case is particularly simple, since the gram matrix has the form $I + c J,$ (where $J$ is the matrix of all ones). $J$ is idempotent, so $(I + c J)^-1 = I + c_1 J,$ by expanding the inverse in a power series (notice that in the spherical case the Gram matrix is positive definite, so $c < 1,$ so we are allowed to do this).
Anyway, when the smoke clears, we see that all of the links of our original simplex $S$ are regular spherical simplices, congruent amongst themselves. The facets of the links correspond to the links of the facets of $S,$ and the result follows by induction (the base case is the two dimensional case which you already know how to deal with).