[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 $X$.
[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]
[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]
