Compact hypersurface of $\mathbb{R}^{n+1}$ with positive curvature is diffeomorphic to $S^n$. I have a compact hypersurface $M$ of $\mathbb{R}^{n+1}$ with positive curvature. I need to show that it is diffeomorphic to $S^n$.
The hint is to consider the shape operator $A_{\nu_p} x$, where $\nu$ is a smooth unit normal vector field regarded as a map $\nu: M \rightarrow S^n$, and then show that it is a covering map.
Unfortunately, I don't think the hint really made the approach any clearer. Can anyone help to shed some light on this exercise for me?

The shape operator is selfadjoint as an endomorphism of the tangent space.  Therefore it can be diagonalized over $\mathbb{R}$. The hypothesis of positive (sectional) curvature then implies that all the eigenvalues are positive. Therefore the determinant is positive and in particular nonzero. It follows that the Gauss map (assigning to each point, the unit normal vector at that point) is a regular map. This is because the shape operator is the tangent map of the Gauss map. But any regular cover of the sphere is the identity because the sphere is simply connected.
