$\omega_{FS}^n$, the top form of Fubini Study metric Given $\mathbb{C}P^n$, we look at the open set $U_0 = \{[Z_0, \cdots, Z_n]: Z_0 \neq 0\}$, and a coordinate map $\phi: U_0 \rightarrow \mathbb{C}^n$ by 
$$[Z_0,\cdots Z_n] \mapsto (Z_1/Z_0, \cdots, Z_n/Z_0).$$
Recall that the Fubini-Study metric $\omega_{FS}$ on $U_0$ is defined as 
$$(\phi^{-1})^*\omega_{FS}  =  \frac{i}{2} \sum_{j,k}\frac{(1+|z|^2)\delta_{jk} - \bar z_j z_k}{(1+|z|^2)^2} dz_j \wedge d\bar z_k.$$
I want to show that 
$$\int_{U_0} \omega_{FS}^n  = \int_{\mathbb{C}^n} (\phi^{-1})^*\omega_{FS}^n = \pi^n$$ 
(I showed this for $\mathbb{C}P^1$) but I am having trouble seeing what the wedge product is for 
$$(\phi^{-1})^*\omega_{FS}^n=  \Big((\phi^{-1})^*\omega_{FS}\Big)^n.$$
Edit:
I did the following calculation for $\mathbb{C}P^2$, we have the wedge product with itself
\begin{gather}
\left(\frac{-z_1 \bar z_2}{(1+|z|^2)^2} dz_2 \wedge d\bar z_1 + \frac{-z_2 \bar z_1}{(1+|z|^2)^2} dz_1 \wedge d\bar z_2 \\ + \frac{((1+|z|^2)-|z_1|^2}{(1+|z|^2)^2} dz_1 \wedge d\bar z_1+\frac{((1+|z|^2)-|z_2|^2}{(1+|z|^2)^2} dz_2 \wedge d\bar z_2\right)^{\wedge 2}\\
= \frac{2}{(1+|z|^2)^3} dz_1\wedge d\bar z_1\wedge dz_2 \wedge d\bar z_2
\end{gather}
Now combine the constants from Fubini-Study metric, we have
\begin{gather}
\int_{U_0} \omega_{FS}^2 = \int_{\mathbb{C}^2} \frac{-1}{4}\frac{2}{(1+|z|^2)^3} dz_1\wedge d\bar z_1\wedge dz_2 \wedge d\bar z_2 \\
= 4\int_{\mathbb{R}^4} \frac{1}{4}\frac{2}{(1+|x|^2)^3} dx_1dx_2dx_3dx_4\\
= 2\left(\frac{2\pi^2}{1!}\right)\int_0^\infty \frac{r^3}{(1+r^2)^3} dr 
\end{gather}
using trig substitution for $[x=\tan \theta]$, I got $\pi^2$ for the final calculation.
 A: Denote 
$$\omega_{FS} = \sum_{i,k=1}^n \frac{i}{2} g_{j\bar k} dz_j \wedge d\bar z_k,$$
we look at the matrix $(dz_j \wedge d\bar z_k)$ for $\mathbb C P^3$.
\begin{bmatrix}
    dz_1\wedge d\bar z_1  \quad     & dz_1\wedge d\bar z_2 \quad& dz_1\wedge d\bar z_3 \\
    dz_2\wedge d\bar z_1    \quad   & dz_2\wedge d\bar z_2 \quad& dz_2\wedge d\bar z_3 \\  
    dz_3\wedge d\bar z_1    \quad   & dz_3\wedge d\bar z_2 \quad& dz_3\wedge d\bar z_3
\end{bmatrix} 
we see all the terms in $\omega_{FS}^3$ corresponds to picking three entries in the above matrix, and a non-zero term in $\omega_{FS}^3$ means the three entries has to come from different roll and column. This pattern is closely related to computing the determinant using Laplace expansion.
In general for $\mathbb C P^n$, we have
$$\boxed{\omega_{FS}^n = \left(\frac{i}{2}\right)^n n! \det(g_{i\bar j}) dz_1\wedge d\bar z_1 \wedge \cdots \wedge dz_n \wedge d\bar z_n.}$$
Recall $g_{j\bar k} = \frac{(1+|z|^2)\delta_{jk} - \bar z_j z_k}{(1+|z|^2)^2}$, from a matrix determinant lemma
$$\det(A+uv^T) = (1+ v^TA^{-1}u)\det(A),$$
we have 
$$\boxed{\det(g_{j\bar{k}}) = \frac{1}{(1+|z|^2)^{n+1}}.}$$
We have 
\begin{align*}
\int_{U_0} \omega_{FS}^n &= \int_{\mathbb C^n} \left(\frac{i}{2}\right)^n n! \det(g_{i\bar j}) dz_1\wedge d\bar z_1 \wedge \cdots \wedge dz_n \wedge d\bar z_n\\
&= \left(\frac{i}{2}\right)^n n! (-2i)^n \text{vol}(S_{2n-1}) \int_0^\infty \frac{r^{2n-1}}{(1+r^2)^{n+1}} dr\\
&=  \pi^n 2n \int_0^{\pi/2} \sin^{2n-1}\theta \cos \theta d\theta\\
&=\pi^n
\end{align*}
Once again, since $U_0^c$ is a null set, we have $\int_{\mathbb C P^n} \omega_{FS}^n = \pi^n$.
