Cartesian powers of shapes of FANR's and polyhedra

We also prove that if (X,x)∈FANR, then Shn(X,x)=Sh(X,x), for some integer n≥2, implies that Sh(X)=1. This resolves positively Problem (5.6) from [7] (1981). Similarly as for polyhedra, first we show that if X∈FANR and Sh(X)≠1, then X cannot have the shape of a proper Cartesian factor of itself.