= Solution
Let $S=k[X_0,\ldots,X_n]$ with its usual grading, and put $S(m)_d=S_{m+d}$. Define the <twisting sheaf on projective space> by
$$
\mathcal O_{\mathbb P^n}(m)=\widetilde{S(m)}.
$$
More explicitly, on $U_i=D_+(X_i)$ its sections are the homogeneous degree-$m$ part of $S_{X_i}$. They are a free rank-one <module> over $(S_{X_i})_0=k[X_j/X_i:j\ne i]$, with generator $e_i=X_i^m$. These generators make sense for every integer $m$, since $X_i$ is invertible on this chart. Their transition rule is
$$
e_i=(X_i/X_j)^m e_j.
$$
The ratios are regular units on overlaps and satisfy the cocycle identity, so they glue to an <invertible sheaf>. In particular $\mathcal O(m)\otimes\mathcal O(r)\cong\mathcal O(m+r)$.
For $n\ge1$, a <global section> is a compatible homogeneous degree-$m$ rational expression lying in every $S_{X_i}$. Inside the <fraction field> of $S$,
$$
\bigcap_{i=0}^nS_{X_i}=S.
$$
To justify this, write a fraction in lowest terms in the <unique factorization domain> $S$. Membership in $S_{X_i}$ forces every irreducible divisor of its denominator to be associated with $X_i$. Since there are at least two nonassociated variables, the denominator must be a unit. Therefore
$$
H^0(\mathbb P^n,\mathcal O(m))=
\begin{cases}
S_m,&m\ge0,\\
0,&m<0,
\end{cases}
\qquad(n\ge1).
$$
The degree-$m$ <monomials> are indexed by nonnegative integers $a_0,\ldots,a_n$ with sum $m$. The <stars and bars> count gives
$$
\boxed{h^0(\mathcal O_{\mathbb P^n}(m))=
\begin{cases}
\binom{m+n}{n},&m\ge0,\\
0,&m<0,
\end{cases}\qquad(n\ge1).}
$$
Here and below, an all-integer binomial shorthand must use the counting convention $\binom{a}{n}=0$ when $a<n$, including negative $a$. The polynomial extension of binomial coefficients to negative upper arguments does not give the stated dimensions. In dimension zero, $\mathbb P^0$ is one point and every $\mathcal O_{\mathbb P^0}(m)$ is trivial, so its space of <global sections> is $k$ for every $m$; this case is treated separately.
For a closed inclusion $i:H\hookrightarrow\mathbb P^n$, <direct image sheaf> formation is exact: its stalk is the original stalk at a point of $H$ and zero outside $H$. It preserves <flabby sheaves>, since restrictions on the target are restrictions between the corresponding open subsets of $H$. Push forward a <flasque resolution> of $\mathcal F$; exactness gives a <flasque resolution> of $i_*\mathcal F$, and its complex of <global sections> is the original one because
$$
\Gamma(\mathbb P^n,i_*\mathcal I^j)=\Gamma(H,\mathcal I^j).
$$
Taking cohomology of these identical complexes proves <sheaf cohomology under a closed inclusion>:
$$
\boxed{H^r(H,\mathcal F)\cong H^r(\mathbb P^n,i_*\mathcal F)\quad(r\ge0).}
$$
This argument applies even without quasi-coherence.
We derive the remaining <sheaf cohomology> dimensions using the given eventual vanishing, the <long exact sequence in sheaf cohomology>, and induction on dimension. Multiplication by the linear equation of a <hyperplane> $H\cong\mathbb P^{n-1}$ gives the <hyperplane exact sequence for twisting sheaves>
$$
0\longrightarrow\mathcal O_{\mathbb P^n}(m-1)
\longrightarrow\mathcal O_{\mathbb P^n}(m)
\longrightarrow i_*\mathcal O_H(m)\longrightarrow0.
$$
Exactness can be checked in each chart by quotienting its <polynomial ring> by the equation of $H$; multiplication is <injective> because that equation is a <non-zero-divisor>. We use the closed-inclusion cohomology identification to replace the cohomology of the final <sheaf> by that on $H$.
The dimension-zero base is a point, whose global-section functor is exact. Thus all its higher <sheaf cohomology> is zero. For $n=1$, the relevant <long exact sequence> is
$$
0\to H^0(\mathcal O(m-1))\to H^0(\mathcal O(m))
\to k\to H^1(\mathcal O(m-1))\to H^1(\mathcal O(m))\to0.
$$
When $m\ge0$, restriction of <homogeneous polynomials> to the hyperplane point is <surjective>: after choosing that point as $[1:0]$, the section $X_0^m$ restricts to a generator. Therefore the consecutive $H^1$ groups are isomorphic for $m\ge0$. The given vanishing for large $m$ propagates down to $m=-1$. For $m<0$, both displayed $H^0$ groups are zero, and the sequence becomes
$$
0\to k\to H^1(\mathcal O(m-1))\to H^1(\mathcal O(m))\to0.
$$
Starting with $H^1(\mathcal O(-1))=0$, induction downward gives $h^1(\mathcal O(m))=-m-1$ for $m\le-2$. For degrees $i>1$, the point contributes no cohomology to the exact sequence, so consecutive $H^i$ groups are isomorphic for every $m$; the given eventual vanishing makes them all zero. Hence the complete positive-degree formula holds for $\mathbb P^1$.
Now let $n\ge2$ and assume the formulas for $H\cong\mathbb P^{n-1}$. For every $m$, the restriction
$$
H^0(\mathbb P^n,\mathcal O(m))\to H^0(H,\mathcal O_H(m))
$$
is <surjective>: for $m\ge0$ extend a polynomial in the hyperplane coordinates to the same polynomial on $\mathbb P^n$, and for $m<0$ both groups vanish because $H$ has positive dimension. Consequently the exact sequence gives an <injection>
$$
H^1(\mathbb P^n,\mathcal O(m-1))
\hookrightarrow H^1(\mathbb P^n,\mathcal O(m)).
$$
For $2\le i\le n-1$, the induction hypothesis says $H^{i-1}(H,\mathcal O_H(m))=0$, and gives the same <injection> in degree $i$. For each fixed $m$, composing finitely many <injections> reaches a sufficiently large twist, whose higher cohomology is zero by the given hypothesis. Therefore
$$
H^i(\mathbb P^n,\mathcal O(m))=0
\quad(0<i<n,\ \text{every }m).
$$
For $i>n$, the two neighbouring hyperplane groups $H^{i-1}(H,\mathcal O_H(m))$ and $H^i(H,\mathcal O_H(m))$ are zero by induction. Consecutive $H^i$ groups on $\mathbb P^n$ are then isomorphic, and eventual vanishing again makes them zero. This derives vanishing above the dimension too, without needing a special projective-space cohomology theorem.
For top cohomology the exact sequence now reduces to
$$
0\to H^{n-1}(H,\mathcal O_H(m))
\to H^n(\mathbb P^n,\mathcal O(m-1))
\to H^n(\mathbb P^n,\mathcal O(m))\to0.
$$
Write $b_n(m)=h^n(\mathcal O_{\mathbb P^n}(m))$. The induction hypothesis and eventual vanishing give the recurrence
$$
b_n(m-1)=b_n(m)+b_{n-1}(m),\qquad
b_{n-1}(m)=
\begin{cases}
\binom{-m-1}{n-1},&m\le-n,\\
0,&m\ge-n+1.
\end{cases}
$$
Consecutive top groups agree for $m\ge-n+1$, so descending from a large twist gives $b_n(m)=0$ for $m\ge-n$. Below this range, iterating the recurrence gives a finite sum:
$$
b_n(m)=\sum_{j=m+1}^{-n}\binom{-j-1}{n-1}
=\sum_{\ell=n-1}^{-m-2}\binom{\ell}{n-1}
=\binom{-m-1}{n}
\qquad(m\le-n-1).
$$
The last equality follows by summing <Pascal's identity>. These short exact sequences also prove that every group whose dimension is being counted is finite-dimensional. Thus, for $n\ge1$,
$$
\boxed{
h^i(\mathcal O_{\mathbb P^n}(m))=0\quad(0<i\ne n),\qquad
h^n(\mathcal O_{\mathbb P^n}(m))=
\begin{cases}
\binom{-m-1}{n},&m\le-n-1,\\
0,&m\ge-n.
\end{cases}}
$$
With the stated counting convention, the latter is the requested single binomial formula. The base $\mathbb P^0$ has $h^0=1$ for all twists and no higher cohomology; it is not governed by the negative-twist $H^0$ rule for positive-dimensional projective space.
Finally calculate the <canonical line bundle of a smooth variety> from its transition functions. On $U_0$ write $t_j=X_j/X_0$ for $1\le j\le n$. On $U_i$, $i>0$, take the coordinates $u_j=X_j/X_i$ in increasing order of $j\ne i$. Thus
$$
u_0=t_i^{-1},\qquad u_j=t_j/t_i\quad(j\ne0,i).
$$
In their wedge product, $du_0=-t_i^{-2}\,dt_i$ and
$du_j=t_i^{-1}dt_j-t_jt_i^{-2}dt_i$. Every term involving a second $dt_i$ vanishes, so reordering the remaining differentials gives
$$
\bigwedge_{j\ne i}du_j
=(-1)^i t_i^{-(n+1)}\,dt_1\wedge\cdots\wedge dt_n.
$$
This computation is valid in every characteristic. Multiply the ordered local generator on $U_i$ by $(-1)^i$. The resulting generators $\eta_i$ satisfy $\eta_i=(X_i/X_0)^{-n-1}\eta_0$ on $U_i\cap U_0$, and hence the corresponding transition rule on every overlap. They are exactly the transitions of $\mathcal O(-n-1)$. Therefore the <canonical bundle of projective space> is
$$
\boxed{\omega_{\mathbb P^n}\cong\mathcal O_{\mathbb P^n}(-n-1).}
$$
For a <line bundle>, <Serre duality> states
$$
H^i(\mathbb P^n,\mathcal O(m))^*
\cong H^{n-i}(\mathbb P^n,\mathcal O(-m-n-1)).
$$
The intermediate groups on both sides vanish. In the two outer degrees, the top group at twist $m$ has the same dimension as the global-section group at twist $-m-n-1$:
$$
h^n(\mathcal O(m))=h^0(\mathcal O(-m-n-1)).
$$
The polynomial degree on the right is nonnegative precisely when $m\le-n-1$, and its dimension is $\binom{-m-1}{n}$. Interchanging $m$ and $-m-n-1$ checks the other outer degree. On $\mathbb P^0$ both sides are one-dimensional. Thus all the formulas are consistent with <Serre duality>, which was used only for this final consistency check, not to prove the cohomology formulas.
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