Let $X$ be a compact Kahler manifold with Kahler form $\omega$. Then constant functions are obviously harmonic, and if $\alpha$ is harmonic then so is $\omega\wedge \alpha$
because this operation commutes with the Laplacian (see Griffiths-Harris page 115).
When $X=\mathbb{C}\mathbb{P}^n$, $\omega^i$is
the sole harmonic form of degree $2i$, because it
generates $H^{2i}(X)$.

**Footnotes**

- Of course, I meant that $\omega^i$ is a basis for degree $2i$ harmonic forms.
- There are no harmonic forms of odd degree, because there is no cohomology.
- Curiously, this argument is valid for any Kahler metric on projective space, not just the Fubini-Study metric.
- We can argue differently by noting the Fubini-Study metric is invariant under the action of $U(n+1)$, so the same is true for harmonic forms...