[proofplan]
The proof has two ingredients. First, positivity of $L^{\otimes m}$ lets Kodaira vanishing kill all higher cohomology of $K_X \otimes L^{\otimes m}$, so the Euler characteristic equals the dimension of global holomorphic sections. Second, Hirzebruch--Riemann--Roch computes that Euler characteristic as an integral of characteristic classes. Expanding the Chern character shows that the unique contribution of order $m^n$ is $\frac{m^n}{n!}\left\langle c_1(L)^n,[X]\right\rangle$, where $[X] \in H_{2n}(X;\mathbb{Z})$ is the fundamental class and $\left\langle \cdot,[X]\right\rangle$ is the cohomology-fundamental-class pairing; all remaining terms have degree at most $n-1$ in $m$.
[/proofplan]
[step:Use Kodaira vanishing to identify the Euler characteristic with global sections]
Fix an integer $m \geq 1$, and define the holomorphic line bundle
\begin{align*}
E_m := K_X \otimes L^{\otimes m}.
\end{align*}
Since $L$ is positive, its tensor power $L^{\otimes m}$ is positive. By the [Kodaira Vanishing Theorem](/page/Kodaira%20Vanishing%20Theorem), applied to the positive holomorphic line bundle $L^{\otimes m} \to X$, we have
\begin{align*}
H^q(X,K_X \otimes L^{\otimes m}) = 0
\end{align*}
for every integer $q>0$. Therefore
\begin{align*}
\chi(X,E_m)
&=
\sum_{q=0}^{n} (-1)^q \dim_{\mathbb{C}} H^q(X,E_m) \\
&=
\dim_{\mathbb{C}} H^0(X,E_m).
\end{align*}
[guided]
Fix an integer $m \geq 1$. The bundle whose holomorphic sections we are counting is
\begin{align*}
E_m := K_X \otimes L^{\otimes m}.
\end{align*}
The goal in this step is to replace the alternating Euler characteristic by the ordinary dimension of $H^0$. This is exactly what vanishing of higher cohomology gives.
Because $L$ is positive, every positive tensor power $L^{\otimes m}$ is again positive. The [Kodaira Vanishing Theorem](/page/Kodaira%20Vanishing%20Theorem) applies to the compact Kähler manifold $X$ and the positive holomorphic line bundle $L^{\otimes m} \to X$. Its conclusion is
\begin{align*}
H^q(X,K_X \otimes L^{\otimes m}) = 0
\end{align*}
for every integer $q>0$. Hence every term in the Euler characteristic except the $q=0$ term vanishes:
\begin{align*}
\chi(X,E_m)
&=
\sum_{q=0}^{n} (-1)^q \dim_{\mathbb{C}} H^q(X,E_m) \\
&=
\dim_{\mathbb{C}} H^0(X,E_m).
\end{align*}
This proves the asserted equality between the dimension of global sections and the holomorphic Euler characteristic.
[/guided]
[/step]
[step:Apply Hirzebruch--Riemann--Roch to the adjoint bundle]
Let $[X] \in H_{2n}(X;\mathbb{Z})$ denote the fundamental class of the underlying oriented real manifold, and let $\left\langle \cdot,[X]\right\rangle: H^{2n}(X;\mathbb{Q}) \to \mathbb{Q}$ denote the canonical cohomology-fundamental-class pairing. Let $\operatorname{ch}$ denote the Chern character map from complex $K$-theory, or equivalently from holomorphic vector bundles, to even rational cohomology. Let $\operatorname{Td}(T^{1,0}X) \in H^{\mathrm{even}}(X;\mathbb{Q})$ denote the Todd class of the holomorphic tangent bundle $T^{1,0}X \to X$. By the [Hirzebruch--Riemann--Roch Theorem](/page/Hirzebruch--Riemann--Roch%20Theorem), applied to the holomorphic vector bundle $E_m \to X$, we have
\begin{align*}
\chi(X,E_m)
=
\left\langle \operatorname{ch}(E_m)\operatorname{Td}(T^{1,0}X),[X]\right\rangle.
\end{align*}
The Chern character is multiplicative under tensor products of line bundles, so
\begin{align*}
\operatorname{ch}(E_m)
&=
\operatorname{ch}(K_X \otimes L^{\otimes m}) \\
&=
\operatorname{ch}(K_X)\operatorname{ch}(L^{\otimes m}).
\end{align*}
For a holomorphic line bundle $M \to X$, the Chern character is $\operatorname{ch}(M)=\exp(c_1(M))$. Since
\begin{align*}
c_1(L^{\otimes m}) = m c_1(L),
\end{align*}
we obtain
\begin{align*}
\operatorname{ch}(L^{\otimes m})=\exp(m c_1(L)).
\end{align*}
Thus
\begin{align*}
\chi(X,E_m)
=
\left\langle \exp(m c_1(L))\operatorname{ch}(K_X)\operatorname{Td}(T^{1,0}X),[X]\right\rangle.
\end{align*}
[guided]
We now compute the Euler characteristic using the [Hirzebruch--Riemann--Roch Theorem](/page/Hirzebruch--Riemann--Roch%20Theorem). Let $[X] \in H_{2n}(X;\mathbb{Z})$ denote the fundamental class of the underlying oriented real manifold, and let $\left\langle \cdot,[X]\right\rangle: H^{2n}(X;\mathbb{Q}) \to \mathbb{Q}$ denote the canonical cohomology-fundamental-class pairing. Let $\operatorname{ch}$ denote the Chern character map from complex $K$-theory, or equivalently from holomorphic vector bundles, to even rational cohomology. Let $\operatorname{Td}(T^{1,0}X) \in H^{\mathrm{even}}(X;\mathbb{Q})$ denote the Todd class of the holomorphic tangent bundle $T^{1,0}X \to X$.
The theorem applies because $X$ is compact complex, hence the holomorphic Euler characteristic of a holomorphic vector bundle is finite, and $E_m \to X$ is a holomorphic line bundle. Hirzebruch--Riemann--Roch gives
\begin{align*}
\chi(X,E_m)
=
\left\langle \operatorname{ch}(E_m)\operatorname{Td}(T^{1,0}X),[X]\right\rangle.
\end{align*}
Next we expand the Chern character of $E_m$. Since
\begin{align*}
E_m = K_X \otimes L^{\otimes m},
\end{align*}
and the Chern character is multiplicative for tensor products of line bundles, we get
\begin{align*}
\operatorname{ch}(E_m)
=
\operatorname{ch}(K_X)\operatorname{ch}(L^{\otimes m}).
\end{align*}
For a holomorphic line bundle $M \to X$, the Chern character is the exponential of the first Chern class:
\begin{align*}
\operatorname{ch}(M)=\exp(c_1(M)).
\end{align*}
Applying this to $M=L^{\otimes m}$ and using additivity of $c_1$ under tensor powers gives
\begin{align*}
c_1(L^{\otimes m}) = m c_1(L),
\end{align*}
so
\begin{align*}
\operatorname{ch}(L^{\otimes m})=\exp(m c_1(L)).
\end{align*}
Substituting into Hirzebruch--Riemann--Roch yields
\begin{align*}
\chi(X,E_m)
=
\left\langle \exp(m c_1(L))\operatorname{ch}(K_X)\operatorname{Td}(T^{1,0}X),[X]\right\rangle.
\end{align*}
This formula isolates all dependence on $m$ inside the exponential factor.
[/guided]
[/step]
[step:Extract the degree $2n$ term and identify the leading coefficient]
Define the fixed cohomology class
\begin{align*}
A := \operatorname{ch}(K_X)\operatorname{Td}(T^{1,0}X) \in H^{\mathrm{even}}(X;\mathbb{Q}).
\end{align*}
Write its homogeneous decomposition as
\begin{align*}
A = A_0 + A_2 + \cdots + A_{2n},
\end{align*}
where $A_{2j} \in H^{2j}(X;\mathbb{Q})$. Since both $\operatorname{ch}(K_X)$ and $\operatorname{Td}(T^{1,0}X)$ have degree-zero component $1$, we have
\begin{align*}
A_0=1.
\end{align*}
The exponential expansion in cohomology is finite in degree at most $2n$:
\begin{align*}
\exp(m c_1(L))
=
\sum_{k=0}^{n} \frac{m^k}{k!}c_1(L)^k.
\end{align*}
Only the degree $2n$ component can be paired with $[X]$. Therefore
\begin{align*}
\chi(X,E_m)
&=
\sum_{k=0}^{n}
\frac{m^k}{k!}
\left\langle c_1(L)^k A_{2n-2k},[X]\right\rangle.
\end{align*}
The coefficient of $m^n$ comes from $k=n$ and $A_0=1$. Every remaining summand has $k \leq n-1$, hence has degree at most $n-1$ as a polynomial in $m$. Therefore the leading term of $\chi(X,E_m)$ is
\begin{align*}
\frac{m^n}{n!}\left\langle c_1(L)^n,[X]\right\rangle.
\end{align*}
[guided]
Let us separate the fixed characteristic-class contribution from the part that depends on $m$. Define
\begin{align*}
A := \operatorname{ch}(K_X)\operatorname{Td}(T^{1,0}X) \in H^{\mathrm{even}}(X;\mathbb{Q}).
\end{align*}
This class does not depend on $m$. Decompose it by cohomological degree:
\begin{align*}
A = A_0 + A_2 + \cdots + A_{2n},
\end{align*}
where $A_{2j} \in H^{2j}(X;\mathbb{Q})$. The degree-zero component is
\begin{align*}
A_0=1,
\end{align*}
because both $\operatorname{ch}(K_X)$ and $\operatorname{Td}(T^{1,0}X)$ have degree-zero component equal to $1$.
Now expand the exponential. Since $X$ has complex dimension $n$, its real dimension is $2n$, and cohomology classes of degree greater than $2n$ do not contribute to integration over $X$. Thus
\begin{align*}
\exp(m c_1(L))
=
\sum_{k=0}^{n} \frac{m^k}{k!}c_1(L)^k.
\end{align*}
Multiplying this expansion by $A$, the only terms that can be paired with $[X]$ are the terms of total cohomological degree $2n$. The factor $c_1(L)^k$ has degree $2k$, so it must be paired with $A_{2n-2k}$. Hence
\begin{align*}
\chi(X,E_m)
&=
\sum_{k=0}^{n}
\frac{m^k}{k!}
\left\langle c_1(L)^k A_{2n-2k},[X]\right\rangle.
\end{align*}
This expression is a polynomial in $m$ of degree at most $n$. The top-degree term occurs when $k=n$. In that case the companion factor is $A_0$, and we have already identified $A_0=1$. Therefore the coefficient of $m^n$ is
\begin{align*}
\frac{1}{n!}\left\langle c_1(L)^n,[X]\right\rangle.
\end{align*}
Every remaining summand has $k \leq n-1$, so every remaining summand is a polynomial term in $m$ of degree at most $n-1$. Consequently, the leading term of $\chi(X,E_m)$ is
\begin{align*}
\frac{m^n}{n!}\left\langle c_1(L)^n,[X]\right\rangle.
\end{align*}
[/guided]
[/step]
[step:Combine vanishing with Riemann--Roch to obtain the asymptotic dimension]
From the first step,
\begin{align*}
\dim_{\mathbb{C}} H^0(X,E_m)=\chi(X,E_m)
\end{align*}
for every integer $m \geq 1$. From the Riemann--Roch expansion, the leading term of $\chi(X,E_m)$ is
\begin{align*}
\frac{m^n}{n!}\left\langle c_1(L)^n,[X]\right\rangle.
\end{align*}
Substituting $E_m=K_X\otimes L^{\otimes m}$ gives that the leading term of
\begin{align*}
\dim_{\mathbb{C}} H^0(X,K_X\otimes L^{\otimes m})
\end{align*}
is
\begin{align*}
\frac{m^n}{n!}\left\langle c_1(L)^n,[X]\right\rangle,
\end{align*}
which is the asserted leading asymptotic term.
[guided]
The first step converted the dimension problem into an Euler characteristic problem. For every integer $m \geq 1$, it proved
\begin{align*}
\dim_{\mathbb{C}} H^0(X,E_m)=\chi(X,E_m),
\end{align*}
where
\begin{align*}
E_m=K_X\otimes L^{\otimes m}.
\end{align*}
The Riemann--Roch computation then identified the leading term of this Euler characteristic as
\begin{align*}
\frac{m^n}{n!}\left\langle c_1(L)^n,[X]\right\rangle.
\end{align*}
Since the two quantities are equal for each $m \geq 1$, they have the same leading term as $m \to \infty$. Substituting the definition of $E_m$ gives that the leading term of
\begin{align*}
\dim_{\mathbb{C}} H^0(X,K_X\otimes L^{\otimes m})
\end{align*}
is
\begin{align*}
\frac{m^n}{n!}\left\langle c_1(L)^n,[X]\right\rangle.
\end{align*}
This completes the proof of both the equality with the Euler characteristic and the asserted asymptotic growth of the dimension of adjoint sections.
[/guided]
[/step]
