Show that $\mathbb{B}^n $ is a smooth manifold with its boundary diffeomorphic to $S^{n-1}$

Consider a closed unit ball $\mathbb{B}^n = \{ x\in\mathbb{R^n} : \|x\|\le 1\} $

How do I show that $\mathbb{B}^n $ is a smooth manifold with its boundary ($\partial \mathbb{B^n}$) diffeomorphic to $S^{n-1}$ ?

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There is a lemma in Milnor’s “Topology from the differentiable viewpoint” (it’s the lemma 3 on the chapter 2) that says:
Let $M$ be a manifold without boundary and $g:M\to \mathbb R$ a smooth map with $0$ as regular value. The set $N=\{x\in M: g(x)\geq 0\}$ is a submanifold of $M$ with boundary and $\partial N = g^{-1}(0)$. Now, take $M=\mathbb R^m$ and $g(x)=1-|x|^2$ and you will obtain
$$N=\{x\in \mathbb R^m: 1-|x|^2\geq 0\} = \mathbb B^m$$
and $$\partial N = \mathbb S^{m-1}.$$