Volume form on a sphere.
We have
$$w = \frac{1}{r} \sum_{i=1}^{n+1} (-1)^{i-1} x_i dx_1 \cdots\hat{dx_i},\cdots dx_{n+1} \ ,$$
thus
$$dw = \frac{1}{r} \sum_{i=1}^{n+1} (-1)^{i-1} dx_i dx_1 \cdots\hat{dx_i},\cdots dx_{n+1} = \frac{1}{r} \sum_{i=1}^{n+1} dx_1\cdots dx_{n+1} = \frac{n+1}{r} dx_1 \cdots dx_{n+1}\ .$$
Note that the second equality holds because in commuting those $dx_j$ we have the rule
$$dx_i dx_j = - dx_j dx_i\ .$$
Now by Stokes theorem, as $\mathbb S^n = \partial B^{n+1}$,
$$\int_{\mathbb S^n} w = \int_{\partial B^{n+1}} w = \int_{\mathbb B^{n+1}} dw = (n+1) V(B^{n+1})\neq 0$$
Thus $w$ when restricted to $\mathbb S^n$ is not exact (that it is not exact in $\mathbb R^{n+1}$ is obvious as $dw \neq 0$). The reason is again by Stokes theorem: if $w = d\alpha$, then
$$\int_{\mathbb S^n} w = \int_{\mathbb S^n} d\alpha = 0\ .$$
But we have seen that this is not zero.
John Ma interpreted the form $w$ differently than I believe was intended. He interpreted $w$ as a constant called $1/r$ times an $n$-form. He then ignored the constant and correctly used Stokes theorem to find the integral of said $n$-form over the sphere. I believe that, in fact, $r$ was intended to be a function of the $x$ variables. In any case, let me add here a computation of $dw$ with this latter interpretation of $w$.
If $w$ depends on $r$ then one gets an additional term in John Ma's formula for $dw$. Indeed, since
$$ r^{-1} = |x|^{-1}= \left(\sum_j x_j^2 \right)^{-1/2} $$
we find that
$$ d r^{-1} = -|x|^{-3} \cdot \sum_j x_j dx_j $$
Thus, the exterior derivative $dw$ has the term indicated in the answer of John Ma plus
$$ -|x|^{-3} \left(\sum x_j dx_j \right) \wedge \left(\sum_i (-1)^{i-1} x_i \alpha_i \right) $$
where $\alpha_i$ denotes the wedge product of all of the $dx_j$'s except for $dx_i$. Since $dx_i \wedge dx_{i+1} = -dx_{i+1} \wedge dx_i$ and $dx_i \wedge dx_i=0$, the last displayed expression simplifies to
$$ -|x|^{-3} \sum_{i,j} (-1)^{i-1} x_jx_i~ dx_j \wedge \alpha_i~ =~ -|x|^{-3}\sum_j x_j^2~dV~ =~ -|x|^{-1} dV$$
where $dV$ is the product of all $dx_i$'s without exception.
So we have
$$ dw~ =~ \frac{n}{r} \cdot dV. $$
Because $w$ depends on $r$, then applying Stokes theorem
$$ \int_{\Omega} d \omega = \int_{\partial \Omega} \omega $$
requires some care. Indeed, $w$ is not defined---as is---on the closed unit ball $B$. In particular, it is not defined at $x=0$. If $n>0$, then by setting $\omega_x \equiv 0$ for $x=0$ we obtain a continuous extension to ${\mathbb R}^{n+1}$. However, this extension is not differentiable if $n=1$. For $n>0$, one can apply Stokes theorem on the domain $B_1-B_{\epsilon}$ where $B_r$ denotes the ball of radius $r$. Then Stokes theorem reads
$$ \int_{B_1-B_{\epsilon}} dw~ = \int_{\partial B_1} w~ +~ \int_{\partial B_{\epsilon}} w $$
where one should pay attention to orientations of the boundary. For $n>0$, $\|\omega_x\|$ tends to zero as $x$ tends to zero, and hence the latter integral tends to zero. Thus we have
$$ \int_{B_1} \frac{n}{r} \cdot dV~ =~ \int_{\partial B_1} w$$
Also if $w$ depends on $r$, then we can use polar coordinates to evaluate the integral. That is, write $dV= r^n dr \wedge d \omega$ where $d \omega$ is the usual measure on the unit sphere. The above integral becomes
$$ \int_{\partial B_1} \int_{0}^1 n \cdot r^{n-1} \cdot dr\, d\omega~ = \int_{\partial B_1} d \omega~ =~ {\rm Vol} (\partial B_1)$$
If $w$ were exact, then $w=d \alpha$ and by Stokes
$$ \int_{\partial B_1} w~ =~ \int_{ \partial B_1} d \alpha =\int_{\partial (\partial B_1)} \alpha$$
But the sphere has no boundary and so the latter integral is zero. This contradicts the fact that the sphere has nonzero volume.
Of course, if you're only interested in showing that the form is not exact. then it's easier not to consider $w$ but rather $r \cdot w$ as John Ma does.