Non-degeneracy of the Fubini-Study form on $\mathbb{C}P^n$

I am trying to solve exercise 1.9.2 from "An Introduction to Differential Geometry" by Vincent Minerbe (link: https://webusers.imj-prg.fr/uploads//vincent.minerbe/Geodiff/m2dg.pdf page 33). In short, we consider $$\mathbb{S}^{2n+1}$$ to be the unit sphere in $$\mathbb{C}^{n+1}$$, and we define a 1-form $$\eta$$ on it, given by $$\eta_z(V)=\langle iz | V\rangle$$ for each $$z\in\mathbb{S}^{2n+1}$$ (here $$\langle\cdot | \cdot\rangle$$ denotes the standard hermitian product on $$\mathbb{C}^{n+1}$$). I managed to prove that there is a unique 2-form $$\Omega$$ on $$\mathbb{C}P^n\cong \mathbb{S}^{2n+1}/\mathbb{S}^1$$ such that $$\pi^*\Omega=d\eta$$, where $$\pi$$ is the quotient map $$\mathbb{S}^{2n+1}\rightarrow \mathbb{C}P^n$$. The exercise then asks to prove that $$\Omega$$ is symplectic (closed and non-degenerate), and the hint is to show that $$\eta\wedge (d\eta)^n$$ is a volume form on $$\mathbb{S}^{2n+1}$$. I am stuck precisely here: I can't show that $$\eta \wedge (d\eta)^n$$ is always non-zero. The idea is to prove that for every point in $$z\in\mathbb{S}^{2n+1}$$ there exists a basis of the tangent space $$(v_1, \dots, v_{2n+1})$$ such that $$\eta\wedge (d\eta)^n_z(v_1, \dots, v_{2n+1})\neq 0$$. Choosing $$v_1=iz$$ and looking for the remaining vectors in the orthogonal (w.r.t. the hermitian metric of $$\mathbb{C}^{n+1}$$) makes the calculations a bit easier, but I still don't manage to compute an expression for $$d\eta$$, not even with these restrictions.

My question is: may I have a hint on how to prove that $$\eta\wedge (d\eta)^n\neq 0$$ without using an explicit formula for $$d\eta$$? If this tasks appears impossible, given an atlas for $$\mathbb{S}^{2n+1}$$ how can I find an expression for the hermitian product between two tangent vectors as a function of their coordinates?

I have no interest in trying to do this in a coordinate-free manner. You should note that, using complex coordinates $$(z_0,\dots,z_n)$$ on $$\Bbb C^{n+1}$$, the $$1$$-form $$\eta$$ is given by $$\eta = \sqrt{-1}\sum z_j\,d\bar z_j,$$ since $$\langle z,V\rangle = \sum z_j\overline V_j = \sum z_j\,d\bar z_j(V)$$. Then $$d\eta = \sqrt{-1}\sum dz_j\wedge d\bar z_j$$, and the rest is easy.
(By the way, if you've never thought about it, since $$z$$ and $$v$$ are real orthogonal, the hermitian inner product has real part $$0$$ and so $$i\langle z,V\rangle$$ is necessarily real.)