[proofplan]
We compute the determinant of the cotangent bundle using the dual [Euler sequence on projective space](/theorems/7023). The sequence identifies $\Omega^1_{\mathbb{P}^n}$ as the kernel of a map from $\mathcal{O}_{\mathbb{P}^n}(-1)^{\oplus(n+1)}$ to $\mathcal{O}_{\mathbb{P}^n}$. Taking determinants in this short exact sequence reduces the computation to the determinant of a direct sum of identical line bundles, which is $\mathcal{O}_{\mathbb{P}^n}(-n-1)$.
[/proofplan]
[step:Apply the dual Euler sequence on projective space]
Set $X:=\mathbb{P}^n_{\mathbb{C}}$. We use the dual [Euler sequence for complex projective space](/theorems/7018), which states that on $\mathbb{P}^n_{\mathbb{C}}$ there is a short exact sequence of locally free sheaves
\begin{align*}
0\longrightarrow \Omega^1_X\longrightarrow \mathcal{O}_X(-1)^{\oplus(n+1)}\longrightarrow \mathcal{O}_X\longrightarrow 0.
\end{align*}
Its hypotheses match the present theorem because $X$ is precisely complex projective $n$-space. Here $\Omega^1_X$ is the cotangent bundle of $X$, $\mathcal{O}_X$ is the structure line bundle, and $\mathcal{O}_X(-1)$ is the dual of the hyperplane twisting sheaf $\mathcal{O}_X(1)$.
[guided]
Let $X:=\mathbb{P}^n_{\mathbb{C}}$. The canonical bundle is defined as the determinant line bundle of the cotangent bundle, so the object we must compute is $\det \Omega^1_X$.
The main input is the dual Euler sequence for complex projective space. This result applies here because the theorem assumes that $X$ is complex projective $n$-space. It gives the short exact sequence of locally free sheaves
\begin{align*}
0\longrightarrow \Omega^1_X\longrightarrow \mathcal{O}_X(-1)^{\oplus(n+1)}\longrightarrow \mathcal{O}_X\longrightarrow 0.
\end{align*}
The middle term is a direct sum of $n+1$ copies of the dual hyperplane twisting sheaf $\mathcal{O}_X(-1)$, and the quotient is the structure line bundle $\mathcal{O}_X$.
Now we take determinants. For any short exact sequence of vector bundles
\begin{align*}
0\longrightarrow A\longrightarrow B\longrightarrow C\longrightarrow 0,
\end{align*}
there is a natural isomorphism
\begin{align*}
\det B\cong \det A\otimes \det C.
\end{align*}
Applying this with $A=\Omega^1_X$, $B=\mathcal{O}_X(-1)^{\oplus(n+1)}$, and $C=\mathcal{O}_X$ gives
\begin{align*}
\det\left(\mathcal{O}_X(-1)^{\oplus(n+1)}\right)\cong \det(\Omega^1_X)\otimes \det(\mathcal{O}_X).
\end{align*}
Since $\mathcal{O}_X$ is a line bundle, its determinant is itself, and as the structure line bundle it is the tensor identity. Therefore
\begin{align*}
\det(\Omega^1_X)\cong \det\left(\mathcal{O}_X(-1)^{\oplus(n+1)}\right).
\end{align*}
It remains to compute the determinant on the right. The determinant of a direct sum is the [tensor product](/page/Tensor%20Product) of the determinants of the summands. Each summand $\mathcal{O}_X(-1)$ has rank one, so its determinant is $\mathcal{O}_X(-1)$. Hence
\begin{align*}
\det\left(\mathcal{O}_X(-1)^{\oplus(n+1)}\right)\cong \mathcal{O}_X(-1)^{\otimes(n+1)}.
\end{align*}
By the tensor rule for twisting sheaves,
\begin{align*}
\mathcal{O}_X(-1)^{\otimes(n+1)}\cong \mathcal{O}_X(-n-1).
\end{align*}
Combining these isomorphisms gives
\begin{align*}
\det(\Omega^1_X)\cong \mathcal{O}_X(-n-1).
\end{align*}
Finally, by the definition $K_X:=\det\Omega^1_X$, this becomes
\begin{align*}
K_{\mathbb{P}^n}\cong \mathcal{O}_{\mathbb{P}^n}(-n-1).
\end{align*}
[/guided]
[/step]
[step:Take determinants in the short exact sequence]
For a short exact sequence of vector bundles
\begin{align*}
0\longrightarrow A\longrightarrow B\longrightarrow C\longrightarrow 0,
\end{align*}
there is a natural isomorphism of determinant line bundles
\begin{align*}
\det B\cong \det A\otimes \det C.
\end{align*}
Applying this determinant formula to the dual Euler sequence with $A=\Omega^1_X$, $B=\mathcal{O}_X(-1)^{\oplus(n+1)}$, and $C=\mathcal{O}_X$ gives
\begin{align*}
\det\left(\mathcal{O}_X(-1)^{\oplus(n+1)}\right)\cong \det(\Omega^1_X)\otimes \det(\mathcal{O}_X).
\end{align*}
Since $\mathcal{O}_X$ is a line bundle and is the tensor identity, $\det(\mathcal{O}_X)\cong \mathcal{O}_X$, so this becomes
\begin{align*}
\det(\Omega^1_X)\cong \det\left(\mathcal{O}_X(-1)^{\oplus(n+1)}\right).
\end{align*}
[/step]
[step:Compute the determinant of the direct sum of twisting sheaves]
Because the determinant of a direct sum is the tensor product of the determinants of the summands, and each line-bundle summand $\mathcal{O}_X(-1)$ has determinant itself, we have
\begin{align*}
\det\left(\mathcal{O}_X(-1)^{\oplus(n+1)}\right)\cong \mathcal{O}_X(-1)^{\otimes(n+1)}.
\end{align*}
By the defining tensor rule for twisting sheaves,
\begin{align*}
\mathcal{O}_X(-1)^{\otimes(n+1)}\cong \mathcal{O}_X(-n-1).
\end{align*}
Therefore
\begin{align*}
\det(\Omega^1_X)\cong \mathcal{O}_X(-n-1).
\end{align*}
[/step]
[step:Identify the determinant of the cotangent bundle with the canonical bundle]
By definition, the canonical bundle of $X$ is
\begin{align*}
K_X:=\det \Omega^1_X.
\end{align*}
Combining this definition with the determinant computation above gives
\begin{align*}
K_{\mathbb{P}^n}\cong \mathcal{O}_{\mathbb{P}^n}(-n-1).
\end{align*}
This is the desired isomorphism.
[/step]
