 Solution

The regular level set theorem says that a regular level of a smooth map from an $m$-manifold to an $r$-manifold is an $(m-r)$-manifold. The corresponding sublevel-set theorem says that $\{F\le c\}$ is a manifold with boundary $F^{-1}(c)$ when $c$ is a regular value of a smooth real-valued function, with the interior $F<c$.

Apply this to $F(x)=|x{|}^2$ on ${\mathbb{R}}^{n+1}$. On $F^{-1}(1)$ its gradient $2x$ is nonzero. Thus $S^n$ is an $n$-manifold and $B^{n+1}$ is an $(n+1)$-manifold with boundary $\boxed{\partial B^{n+1}=S^n}$. Locally the implicit function theorem supplies $F-1$ as one coordinate; the ball occupies its nonpositive side, which is a half-space after reversing that coordinate. This also covers $n=0$: the sphere consists of two points and the ball is the interval with those endpoints.