Let $\omega$ be the Fubini-Study-Form on complex projective space $\mathbb{P}^N$, normalized s.t. $\int_{\mathbb{P}^1}\omega =1$, and let $1\leq n\leq N$. Then $\int_{\mathbb{P}^n}\omega^n =1$. I've seen a differential geometry proof for that.
But I want to ask about the connection to de Rham cohomology. Is it true, that \begin{align*} \int_{\mathbb{P}^1}\omega =1 \Leftrightarrow \int_{\mathbb{P}^n}\omega^n =1 \Leftrightarrow \omega \text{ generator of } H^2\left(\mathbb{P}^N,\mathbb{Z}\right)? \end{align*}
My ideas so far: (using Mumford's Algebraic Geometry I as a reference for de Rham cohomology)
The $2n$-form $\omega^{n}$ is a basis of $H^{2n}\left(\mathbb{P}^N,\mathbb{Z}\right)$. For the homology, $H_{2n}\left(\mathbb{P}^N,\mathbb{Z}\right)\cong \mathbb{Z}$ and is generated by the fundamental class $[\mathbb{P}^n]$ (the image of $\mathbb{P}^n$ under the homomorphism $H_{2n}\left(\mathbb{P}^n,\mathbb{Z}\right)\rightarrow H_{2n}\left(\mathbb{P}^N,\mathbb{Z}\right)$).
Maybe because we have the bilinear map \begin{align*} H^{2n}\left(\mathbb{P}^N,\mathbb{Z}\right) \times H_{2n}\left(\mathbb{P}^N,\mathbb{Z}\right) \rightarrow \mathbb{R} \end{align*} the basis element of the cohomology space $\omega^n$ is mapped to a generator/ basis element of the image and the image is $\mathbb{Z}$? Maybe we can use the duality of the two spaces in some sense? I also tried using de Rham's Theorem, but it only made sense for a topdimensional form, which is not of interest here.