I'm trying to give a proof of the Hirzebruch signature theorem from a differentiable viewpoint (i.e. using de Rham cohomology). The proof works, except for one small step.
We know that $H^*(\mathbb{C} P^k)=\mathbb{R}[x]/x^{k+1}$ with $x$ the generator of $H^2(\mathbb{C}P^k)$, or more specifically I'm assuming that $x=-e(S)$, i.e. it is given by minus the Euler class of the universal subbundle over $\mathbb{C}P^n$. Now with $x$ defined as such I need to show that $$\int_{\mathbb CP^k}x^k = 1$$ I need this identity to compute the $L$-genus of the complex projective space. My guess is that you would first prove that $x$ is Poincare dual to $\mathbb CP^{k-1}\subset \mathbb CP^k$ so that, $$\int_{\mathbb CP^k}x^k = \int_{\mathbb CP^{k-1}}x^{k-1}$$ and then apply induction.
The integral over the chern class of the tangent bundle equals the euler characteristic of $M$. The Chern class is $(1+x)^{k+1}$, where $x$ is the generator of the dual of the tautological line bundle. So $\int {k+1 \choose k} x^k=k+1$. Hence the integral of $x^k$ must be equal to one, as the euler characteristic of $\mathbb{C} P^k$ is k+1.