[proofplan] We use the holomorphic Euler sequence on complex [projective space](/page/Projective%20Space) to express $T^{1,0}\mathbb{CP}^n$ as the quotient of $\mathcal O(1)^{\oplus(n+1)}$ by the trivial line bundle. The Whitney product formula then converts the short exact sequence into a multiplicative identity for total Chern classes. Finally, the trivial bundle has total Chern class $1$, the hyperplane bundle has total Chern class $1+h$, and the relation $h^{n+1}=0$ in $H^*(\mathbb{CP}^n;\mathbb Z)$ gives the stated equality in the [cohomology ring](/theorems/2271). [/proofplan]

[step:Apply the holomorphic Euler sequence on $\mathbb{CP}^n$]Let \begin{align*} X:=\mathbb{CP}^n \end{align*} and let $\mathcal O_X$ denote the product holomorphic line bundle over $X$. We use the holomorphic [Euler sequence for complex projective space](/theorems/7018): \begin{align*} 0\longrightarrow \mathcal O_X\longrightarrow \mathcal O(1)^{\oplus(n+1)}\longrightarrow T^{1,0}X\longrightarrow 0. \end{align*} Here $\mathcal O(1)^{\oplus(n+1)}$ denotes the [direct sum](/page/Direct%20Sum) of $n+1$ copies of the hyperplane line bundle. This is a short exact sequence of holomorphic complex vector bundles on $X$.[/step]

[guided]Set \begin{align*} X:=\mathbb{CP}^n. \end{align*} The geometric input is the holomorphic Euler sequence for complex projective space: \begin{align*} 0\longrightarrow \mathcal O_X\longrightarrow \mathcal O(1)^{\oplus(n+1)}\longrightarrow T^{1,0}X\longrightarrow 0. \end{align*} In this sequence, $\mathcal O_X$ is the product holomorphic line bundle over $X$, $\mathcal O(1)^{\oplus(n+1)}$ is the direct sum of $n+1$ copies of the hyperplane line bundle, and $T^{1,0}X$ is the holomorphic tangent bundle. Exactness means that the image of $\mathcal O_X$ inside $\mathcal O(1)^{\oplus(n+1)}$ is exactly the kernel of the surjection onto $T^{1,0}X$. Thus the tangent bundle is the quotient bundle \begin{align*} T^{1,0}X\cong \mathcal O(1)^{\oplus(n+1)}/\mathcal O_X. \end{align*} This is the point of using the Euler sequence: it replaces the tangent bundle, whose Chern classes are not immediate from the definition, by a short exact sequence built from line bundles whose Chern classes are known.[/guided]

[step:Use the Whitney product formula for the short exact sequence] By the [Whitney product formula for Chern classes](/theorems/7052) applied to the short exact sequence above, one has \begin{align*} c(\mathcal O(1)^{\oplus(n+1)})=c(\mathcal O_X)c(T^{1,0}X). \end{align*} The hypotheses of the formula are satisfied because all three terms in the Euler sequence are holomorphic complex vector bundles over the same base space $X$, and the sequence is short exact. [/step]

[step:Compute the total Chern class of the direct sum] Since $\mathcal O_X$ is the product complex line bundle, its total Chern class is \begin{align*} c(\mathcal O_X)=1. \end{align*} By definition of $h$, the total Chern class of the hyperplane line bundle is \begin{align*} c(\mathcal O(1))=1+c_1(\mathcal O(1))=1+h. \end{align*} Applying the Whitney product formula repeatedly to the direct sum of $n+1$ copies of $\mathcal O(1)$ gives \begin{align*} c(\mathcal O(1)^{\oplus(n+1)})=c(\mathcal O(1))^{n+1}=(1+h)^{n+1}. \end{align*} Combining this with the identity from the Euler sequence gives \begin{align*} (1+h)^{n+1}=1\cdot c(T^{1,0}X)=c(T^{1,0}X). \end{align*} [/step]

[step:Interpret the formula in the cohomology ring of projective space] The equality is an equality in the cohomology ring \begin{align*} H^*(X;\mathbb Z)\cong \mathbb Z[h]/(h^{n+1}). \end{align*} Thus the formal binomial expression $(1+h)^{n+1}$ is interpreted modulo the relation $h^{n+1}=0$. Therefore it has no nonzero component above degree $2n$, which is consistent with the fact that $T^{1,0}\mathbb{CP}^n$ has complex rank $n$. Since $X=\mathbb{CP}^n$, this proves \begin{align*} c(T^{1,0}\mathbb{CP}^n)=(1+h)^{n+1} \end{align*} in $H^*(\mathbb{CP}^n;\mathbb Z)\cong \mathbb Z[h]/(h^{n+1})$. [/step]