In a nutshell, we show that polynomials and nested polytopes are topological, algebraic and algorithmically equivalent. Given two polytops $A\subseteq B$ and a number $k$, the Nested Polytope Problem (NPP) asks, if there exists a polytope $X$ on $k$ vertices such that $A\subseteq X \subseteq B$. The polytope $A$ is given by a set of vertices and the polytope $B$ is given by the defining hyperplanes. We show a universality theorem for NPP. Given an instance $I$ of the NPP, we define the solutions set of $I$ as $$ V'(I) = \{(x_1,\ldots,x_k)\in \mathbb{R}^{k\cdot n} : A\subseteq \text{conv}(x_1,\ldots,x_k) \subseteq B\}.$$ As there are many symmetries, induced by permutations of the vertices, we will consider the \emph{normalized} solution space $V(I)$. Let $F$ be a finite set of polynomials, with bounded solution space. Then there is an instance $I$ of the NPP, which has a rationally-equivalent normalized solution space $V(I)$. Two sets $V$ and $W$ are rationally equivalent if there exists a homeomorphism $f : V \rightarrow W$ such that both $f$ and $f^{-1}$ are given by rational functions. A function $f:V\rightarrow W$ is a homeomorphism, if it is continuous, invertible and its inverse is continuous as well. As a corollary, we show that NPP is $\exists \mathbb{R}$-complete. This implies that unless $\exists \mathbb{R} =$ NP, the NPP is not contained in the complexity class NP. Note that those results already follow from a recent paper by Shitov. Our proof is geometric and arguably easier.