Finding an $n$-differential form in $\Omega^n \mathbb{R}^{n+1}$ that's never zero on $\mathbb{S}^n$.

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Iv'e been struggling with defining an n-form in $\mathbb{R}^{n+1}$ that's never zero on $\mathbb{S}^n$

I know how to define an $n$-form that's never zero in $\mathbb{S}^n$ by defining a function for each map in a finite atlas and summing these functions up, but defining it in $\mathbb{R}^{n+1}$ is problematic.
I also know that i need to define a function(diffeomorphism) from $\mathbb{S}^n$ to a representation of $\mathbb{S}^n$ in $\mathbb{R}^{n+1}$ $$i: \mathbb{S}^n \longrightarrow \mathbb{R}^{n+1} $$ and for a differential form defined on $\Omega^n \mathbb{R}^{n+1}$, $\omega$, $i^*(ω)$ is defined in $\Omega^n \mathbb{R}^{n+1}$

but i dont really have the idea of combining these two concepts together

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I'm not sure if interior products are something you've learned, but they can help here.

On $\mathbb{R}^{n+1}$, let: \begin{align*} X &= x^1 \frac{\partial}{\partial x^1} +x^2 \frac{\partial}{\partial x^2} +\dotsm+ x^{n+1} \frac{\partial}{\partial x^{n+1}} \\ \nu &= dx^1 \wedge dx^2 \wedge \dotsm \wedge dx^{n+1} \end{align*} Then the $n$-form $\omega =\iota_X\nu$ is nonzero on $S^{n}$.

Intuitively speaking, $\nu$ is the volume form on $\mathbb{R}^n$ which takes the determinant of the matrix of the components of the $n+1$ vectors. The vector field $X$ is normal to $S^n$. Given $p \in S^n$ and linearly independent tangent vectors $v_1,\dots,v_n$ to $S^n$ at $p$, $\omega(v_1,\dots,v_n) = \nu(X_p,v_1,\dots,v_n)$. Since $X_p,v_1,\dots,v_n$ are linearly independent, $\nu$ of them is nonzero.

You can use the fact that $\iota_X$ is an antiderivation to calculate it explicitly. If you try $n=1$ and $n=2$ you'll see what's going on.