= Solution
On the standard <affine charts> $U_i=\{X_i\ne0\}$, use the formal frame $e_i=X_i^m$ and glue rank-one free <sheaves> by
$$
e_j=(X_j/X_i)^m e_i.
$$
The transition functions are regular units on overlaps and satisfy the <cocycle> identity. This constructs the <twisting sheaf on projective space> $\mathcal O(m)$ as an <invertible sheaf> for every integer $m$, including negative $m$.
On a nonempty $U$, a section has local expressions $h_i e_i$, where $h_i\in\mathcal O(U\cap U_i)$. Pulling to the punctured affine cone gives $h_i(X/X_i)X_i^m$. Their agreement on overlaps makes them one homogeneous rational function $r$ of degree $m$, regular on $\pi^{-1}U$. To make its <polynomial> representation explicit, write one nonzero local coefficient as $p(y)/q(y)$ in the affine coordinates. Homogenize $p,q$ to degrees $a,b$; the resulting rational expression is
$$
r=X_i^{m-a+b}\frac{p^{\mathrm{hom}}}{q^{\mathrm{hom}}}.
$$
If the exponent of $X_i$ is negative, move its power to the denominator. Cancel common factors; the greatest common divisor of <homogeneous polynomials> can be chosen homogeneous. Thus $r=F/G$ with $F,G$ coprime <homogeneous polynomials>, $G\ne0$, and $\deg F-\deg G=m$.
Conversely, for such a homogeneous rational function regular on $\pi^{-1}U$, its coefficient $r/X_i^m$ is regular on $U\cap U_i$: restrict to the slice $X_i=1$. These coefficients obey the displayed transition rule and hence define a section. This proves the <homogeneous rational sections of a twisting sheaf> description in both directions, for arbitrary open $U$ rather than only a standard chart.
Now lift a regular degree-zero rational function $f=P/Q$ to the punctured cone. The quotient rule gives
$$
\frac{\partial f}{\partial X_i}=\frac{Q\,\partial_iP-P\,\partial_iQ}{Q^2}.
$$
This is a rational function homogeneous of degree $-1$. It is independent of the chosen representation because differentiation is a <derivation> of the rational <function field>. It is regular wherever $f$ is regular: a derivation of a <polynomial> ring extends to each <localization> by the quotient rule, and <regular functions> are locally such fractions. The preceding homogeneous-section description therefore proves
$$
\boxed{\partial_i f\in\Gamma(U,\mathcal O(-1))}.
$$
Zeros of a particular displayed denominator do not invalidate this argument; regularity is a property of the rational function, which can have another local representation.
The map $f\mapsto(\partial_0f,\ldots,\partial_nf)$ is a $k$-linear <sheaf> derivation with values in $\mathcal O(-1)^{\oplus(n+1)}$. The <universal property of Kähler differentials> consequently defines
$$
D:\Omega^1_{\mathbb P^n}\longrightarrow\mathcal O(-1)^{\oplus(n+1)},\qquad
df\longmapsto(\partial_i f)_i.
$$
Multiplication by the coordinate section $X_i\in\Gamma(\mathbb P^n,\mathcal O(1))$ defines the other map
$$
\sigma:\mathcal O(-1)^{\oplus(n+1)}\longrightarrow\mathcal O,\qquad
(g_i)_i\longmapsto\sum_iX_i g_i.
$$
For a degree-$d$ homogeneous polynomial, differentiating each monomial gives $\sum_iX_i\partial_iP=dP$. Applying this to the equal-degree numerator and denominator gives $\sum_iX_i\partial_i f=0$ for degree-zero $f$, so $\sigma D=0$. This identity is valid in every characteristic.
To prove all the exactness assertions, fix $U_i$ and put $y_j=X_j/X_i$ for $j\ne i$. The <Kähler differential sheaf> is free there on $dy_j$, because a derivation of the polynomial ring is determined freely by its values on the coordinates. Trivialize each $\mathcal O(-1)$ with frame $X_i^{-1}$. In this frame,
$$
D(dy_j)=e_j-y_je_i,\qquad
\sigma((b_0,\ldots,b_n))=b_i+\sum_{j\ne i}y_jb_j,
$$
where $e_j$ now denotes the $j$th coordinate <vector> of the direct sum. The map $\sigma$ is surjective, since its $i$th coefficient is one. Its kernel consists exactly of tuples with $b_i=-\sum_{j\ne i}y_jb_j$, so the $n$ displayed <vectors> $e_j-y_je_i$ form a free <basis> of that kernel. The map $D$ sends the differential <basis> bijectively to this kernel <basis> and is therefore injective. Exactness on these charts proves the <cotangent Euler sequence in homogeneous coordinates>
$$
\boxed{0\to\Omega^1_{\mathbb P^n}\xrightarrow{D}
\mathcal O(-1)^{\oplus(n+1)}\xrightarrow{\sigma}\mathcal O\to0}.
$$
The proof uses no division by an integer and hence no characteristic-zero assumption.
For $n\ge1$, the <cohomology of twisting sheaves on projective space> gives
$$
h^q(\mathbb P^n,\mathcal O(m))=
\begin{cases}
\binom{m+n}{n},&q=0,\ m\ge0,\\
\binom{-m-1}{n},&q=n,\ m\le-n-1,\\
0,&\text{otherwise}.
\end{cases}
$$
In particular, $H^0(\mathcal O)=k$, $H^{q>0}(\mathcal O)=0$, and $H^q(\mathcal O(-1))=0$ for every $q$. The <long exact sequence in sheaf cohomology> of the displayed <Euler sequence> begins
$$
0\to H^0(\Omega^1)\to0\to k\to H^1(\Omega^1)\to0,
$$
and gives zero in the remaining degrees. Thus
$$
\boxed{\dim_k H^q(\mathbb P^n,\Omega^1_{\mathbb P^n})=
\begin{cases}1,&q=1,\\0,&q\ne1,\end{cases}\qquad(0\le q\le n,\ n\ge1)}.
$$
The nonzero group is generated by the connecting image of the section $1$ of $\mathcal O$. If $n=0$ is admitted, $\mathbb P^0$ is a point and its cotangent <sheaf> is zero, so the sole requested group $H^0$ is zero.
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