= Solution
Use <homogeneous coordinates> $X_0,\ldots,X_n$. Multiplication by these degree-one sections defines the <sheaf morphism>
$$
\epsilon:\mathcal O(-1)^{\oplus(n+1)}\longrightarrow\mathcal O,\qquad(s_0,\ldots,s_n)\longmapsto\sum_{j=0}^n X_j s_j.
$$
On $U_i=D_+(X_i)$, write $t_j^{(i)}=X_j/X_i$, with $t_i^{(i)}=1$, and let $E_j^{(i)}$ be the standard vector in component $j$ using the local <line bundle> frame $X_i^{-1}$ of $\mathcal O(-1)$. Then $\epsilon(E_j^{(i)})=t_j^{(i)}$, so $\epsilon$ is surjective since $\epsilon(E_i^{(i)})=1$. Its <kernel> is a <locally free sheaf> with basis
$$
F_j^{(i)}=E_j^{(i)}-t_j^{(i)}E_i^{(i)}\qquad(j\ne i).
$$
Indeed any vector in the kernel is uniquely a sum of these vectors, by solving for its $i$-th component.
The <Kähler differential sheaf> $\Omega^1_{\mathbf P^n/k}$ has basis $dt_j^{(i)}$ on $U_i$, since this chart is a polynomial <affine scheme>. Define the local map $dt_j^{(i)}\mapsto F_j^{(i)}$. To check that these isomorphisms glue, on $U_i\cap U_\ell$ put $u=t_i^{(\ell)}$ and $v=t_j^{(\ell)}$. Then
$$
t_j^{(i)}=\frac vu,\qquad dt_j^{(i)}=u^{-1}dt_j^{(\ell)}-vu^{-2}dt_i^{(\ell)}.
$$
The <twisting sheaf on projective space> frame changes by $E_j^{(i)}=u^{-1}E_j^{(\ell)}$, hence
$$
F_j^{(i)}=u^{-1}F_j^{(\ell)}-vu^{-2}F_i^{(\ell)}.
$$
For $j=\ell$, use $t_\ell^{(\ell)}=1$, $dt_\ell^{(\ell)}=0$ and $F_\ell^{(\ell)}=0$; the same formula applies. Thus the differential and kernel frames have identical transition matrices. We obtain the cotangent form of the <Euler sequence>:
$$
\boxed{0\longrightarrow\Omega^1_{\mathbf P^n/k}\longrightarrow\mathcal O(-1)^{\oplus(n+1)}\xrightarrow{\epsilon}\mathcal O\longrightarrow0.}
$$
This construction works over any <field>, including positive characteristic.
The <Kähler differential sheaf> has rank $n$. Taking <determinant line bundles> in this locally split <short exact sequence of sheaves> gives
$$
\det(\mathcal O(-1)^{\oplus(n+1)})\cong\det(\Omega^1_{\mathbf P^n/k})\otimes\det(\mathcal O).
$$
The left side is the tensor product of $n+1$ copies of $\mathcal O(-1)$, and $\det(\mathcal O)=\mathcal O$. Consequently
$$
\boxed{\bigwedge^n\Omega^1_{\mathbf P^n/k}\cong\mathcal O(-n-1).}
$$
It is the <canonical bundle of projective space>. The determinant identity can also be seen directly by taking the wedge of the kernel basis followed by a lift of the quotient basis; changing that lift by a kernel vector leaves the wedge unchanged.
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