[proofplan]
We apply the Hirzebruch-Riemann-Roch theorem for holomorphic vector bundles to the rank-one bundle $L\to X$. The theorem identifies the holomorphic Euler characteristic with the integral over $X$ of the Chern character of the bundle multiplied by the Todd class of the holomorphic tangent bundle. We then compute the Chern character of the tensor power $L^m$ from the first Chern class and extract the top-degree component to see that the resulting integral is a polynomial in $m$ of degree at most $n$.
[/proofplan]
[step:Apply Hirzebruch-Riemann-Roch to the holomorphic line bundle $L$]
Let $[X]\in H_{2n}(X;\mathbb{Z})$ denote the fundamental class determined by the complex orientation of $X$, and write $\int_X \alpha\, d[X]$ for evaluation of a degree-$2n$ cohomology class $\alpha$ on $[X]$. Let $H^q(X,L)$ denote the $q$-th sheaf cohomology group of the sheaf of holomorphic sections of $L$. Define the holomorphic Euler characteristic of $L$ by
\begin{align*}
\chi(X,L) := \sum_{q=0}^{n} (-1)^q \dim_{\mathbb{C}} H^q(X,L).
\end{align*}
Since $X$ is compact and complex and $L\to X$ is a holomorphic vector bundle of rank $1$, the hypotheses of the [Hirzebruch-Riemann-Roch theorem](/page/Hirzebruch-Riemann-Roch%20Theorem) for holomorphic vector bundles are satisfied with $E := L$. That theorem gives
\begin{align*}
\chi(X,L) = \int_X \operatorname{ch}(L)\operatorname{td}(T^{1,0}X)\, d[X].
\end{align*}
This is the first asserted formula.
[guided]
The relevant global invariant is the holomorphic Euler characteristic. Let $[X]\in H_{2n}(X;\mathbb{Z})$ denote the fundamental class determined by the complex orientation of $X$, and write $\int_X \alpha\, d[X]$ for evaluation of a degree-$2n$ cohomology class $\alpha$ on $[X]$. For the holomorphic line bundle $L\to X$, define
\begin{align*}
\chi(X,L) := \sum_{q=0}^{n} (-1)^q \dim_{\mathbb{C}} H^q(X,L),
\end{align*}
where $H^q(X,L)$ is the $q$-th sheaf cohomology group of the sheaf of holomorphic sections of $L$. Compactness of $X$ ensures these cohomology groups are finite-dimensional by the finiteness theorem for coherent analytic sheaves, so the alternating sum is defined.
Now apply the [Hirzebruch-Riemann-Roch theorem](/page/Hirzebruch-Riemann-Roch%20Theorem) for holomorphic vector bundles. Its input is a compact complex manifold and a holomorphic vector bundle on it. Here $X$ is compact complex by hypothesis, and $L\to X$ is a holomorphic vector bundle because a holomorphic line bundle is a holomorphic vector bundle of rank $1$. Therefore the theorem applies with $E := L$ and yields
\begin{align*}
\chi(X,L) = \int_X \operatorname{ch}(L)\operatorname{td}(T^{1,0}X)\, d[X].
\end{align*}
This proves the first displayed identity in the theorem statement.
[/guided]
[/step]
[step:Compute the Chern character of the tensor powers of $L$]
For $m\in\mathbb{Z}$, define the holomorphic line bundle $L^m\to X$ by $L^m := L^{\otimes m}$ when $m\geq 1$, by $L^0 := X\times\mathbb{C}$ with its product holomorphic line bundle structure, and by $L^m := (L^*)^{\otimes(-m)}$ when $m<0$. The dual $L^*\to X$ is a holomorphic line bundle, and tensor products of holomorphic line bundles are holomorphic line bundles, so $L^m\to X$ is holomorphic for every $m\in\mathbb{Z}$.
The first Chern class is additive under [tensor product](/page/Tensor%20Product) of holomorphic line bundles and satisfies $c_1(L^*)=-c_1(L)$. Hence
\begin{align*}
c_1(L^m) = m c_1(L)
\end{align*}
for every $m\in\mathbb{Z}$. For a holomorphic line bundle $M\to X$, the Chern character is
\begin{align*}
\operatorname{ch}(M) = e^{c_1(M)}.
\end{align*}
Applying this with $M := L^m$ gives
\begin{align*}
\operatorname{ch}(L^m) = e^{c_1(L^m)} = e^{m c_1(L)}.
\end{align*}
Applying the previous step to the holomorphic line bundle $L^m\to X$ gives
\begin{align*}
\chi(X,L^m)=\int_X e^{m c_1(L)}\operatorname{td}(T^{1,0}X)\, d[X].
\end{align*}
[guided]
We now repeat the same Riemann-Roch formula for every integer tensor power of $L$. For $m\in\mathbb{Z}$, define $L^m\to X$ by
\begin{align*}
L^m :=
\begin{cases}
L^{\otimes m}, & m\geq 1,\\
X\times\mathbb{C}, & m=0,\\
(L^*)^{\otimes(-m)}, & m<0.
\end{cases}
\end{align*}
The case $m<0$ is the point that must be checked: the dual $L^*\to X$ is a holomorphic line bundle, and tensor products of holomorphic line bundles are holomorphic line bundles. Hence $L^m\to X$ is holomorphic for every $m\in\mathbb{Z}$.
The first Chern class is additive under tensor product of holomorphic line bundles, and dualization changes the sign of the first Chern class:
\begin{align*}
c_1(L^*)=-c_1(L).
\end{align*}
Therefore the three cases give one formula:
\begin{align*}
c_1(L^m)=m c_1(L)
\end{align*}
for every $m\in\mathbb{Z}$. For a holomorphic line bundle $M\to X$, the Chern character is the exponential of the first Chern class:
\begin{align*}
\operatorname{ch}(M) = e^{c_1(M)}.
\end{align*}
Taking $M := L^m$ gives
\begin{align*}
\operatorname{ch}(L^m) = e^{c_1(L^m)} = e^{m c_1(L)}.
\end{align*}
Since $X$ is compact complex and $L^m\to X$ is holomorphic, the Hirzebruch-Riemann-Roch theorem applies to $L^m$. Substituting the computed Chern character into the Riemann-Roch formula gives
\begin{align*}
\chi(X,L^m)=\int_X e^{m c_1(L)}\operatorname{td}(T^{1,0}X)\, d[X].
\end{align*}
This is the second asserted formula.
[/guided]
[/step]
[step:Extract the top-degree component to obtain a polynomial in $m$]
Let $\operatorname{td}_j(T^{1,0}X)\in H^{2j}(X;\mathbb{Q})$ denote the degree-$2j$ component of $\operatorname{td}(T^{1,0}X)$. Since $X$ has complex dimension $n$, only cohomological degree $2n$ contributes to integration over $X$. Thus
\begin{align*}
\chi(X,L^m)
&= \int_X \left(\sum_{k=0}^{n} \frac{m^k c_1(L)^k}{k!}\right)\left(\sum_{j=0}^{n} \operatorname{td}_j(T^{1,0}X)\right)\, d[X] \\
&= \sum_{k=0}^{n} m^k \int_X \frac{c_1(L)^k}{k!}\operatorname{td}_{n-k}(T^{1,0}X)\, d[X].
\end{align*}
For each $k\in\{0,\dots,n\}$, define
\begin{align*}
a_k := \int_X \frac{c_1(L)^k}{k!}\operatorname{td}_{n-k}(T^{1,0}X)\, d[X] \in \mathbb{Q}.
\end{align*}
Then
\begin{align*}
\chi(X,L^m)=\sum_{k=0}^{n} a_k m^k,
\end{align*}
so $\chi(X,L^m)$ is a polynomial in $m$ of degree at most $n$.
[guided]
The integral over a compact complex manifold of complex dimension $n$ only sees cohomology classes of real degree $2n$. We therefore decompose the Todd class into its homogeneous components. For each $j\in\{0,\dots,n\}$, let
\begin{align*}
\operatorname{td}_j(T^{1,0}X)\in H^{2j}(X;\mathbb{Q})
\end{align*}
denote the degree-$2j$ component of $\operatorname{td}(T^{1,0}X)$.
The exponential $e^{m c_1(L)}$ is interpreted as its finite cohomological expansion on $X$. Terms with $k>n$ have degree greater than $2n$ and do not contribute to the integral. Hence
\begin{align*}
e^{m c_1(L)} = \sum_{k=0}^{n} \frac{m^k c_1(L)^k}{k!}
\end{align*}
for the purpose of integration over $X$. Multiplying by the Todd class and retaining only total degree $2n$ gives
\begin{align*}
\chi(X,L^m)
&= \int_X \left(\sum_{k=0}^{n} \frac{m^k c_1(L)^k}{k!}\right)\left(\sum_{j=0}^{n} \operatorname{td}_j(T^{1,0}X)\right)\, d[X] \\
&= \sum_{k=0}^{n} m^k \int_X \frac{c_1(L)^k}{k!}\operatorname{td}_{n-k}(T^{1,0}X)\, d[X].
\end{align*}
The second equality is the degree-counting step: the product $c_1(L)^k\operatorname{td}_j(T^{1,0}X)$ has real cohomological degree $2k+2j$, so it contributes to integration over $X$ exactly when $k+j=n$.
For each $k\in\{0,\dots,n\}$, define the coefficient
\begin{align*}
a_k := \int_X \frac{c_1(L)^k}{k!}\operatorname{td}_{n-k}(T^{1,0}X)\, d[X] \in \mathbb{Q}.
\end{align*}
Then the formula becomes
\begin{align*}
\chi(X,L^m)=\sum_{k=0}^{n} a_k m^k.
\end{align*}
This is a polynomial in $m$ of degree at most $n$, completing the proof.
[/guided]
[/step]
