Given positive integers $k, m, n$, satisfying $1 \leqslant k \leqslant m \leqslant n$. Try to find
$$
\sum_{i=0}^{n}(-1)^{i} \frac{1}{n+k+i} \cdot \frac{(m+n+i)!}{i!(n-i)!(m+i)!}
$$

and write down the derivation process.
(Xu Yichao, provided)
II. The answer to this question is 0. Below, we will prove this combinatorial identity by constructing polynomials and using interpolation and difference methods.
Method 1: Construct the polynomial
$$
\begin{aligned}
f(x) & =\sum_{i=0}^{n} a_{i} x(x+1) \cdots(x+i-1)(x+i+1) \\
& \cdots(x+n)-(x-m-1) \cdots(x-m-n) .
\end{aligned}
$$

We will appropriately choose the coefficients $a_{i}(0 \leqslant i \leqslant n)$ such that
$$
f(x) \equiv 0 .
$$

Notice that $f(x)$ is a polynomial of degree at most $n$, so $f(x) \equiv 0$ if and only if $f(x)$ has $n+1$ roots $0, -1, -2, \cdots, -n$. Therefore, for $0 \leqslant i \leqslant n$, we should have
$$
\begin{aligned}
0= & f(-i) \\
= & a_{i}(-i)(-i+1) \cdots(-i+i-1)(-i+i+ \\
& 1) \cdots(-i+n)-(-i-m-1) \cdots(-i \\
& -m-n) \cdot \\
= & (-1)^{i} i!(n-i)!a_{i}-(-1)^{n} \\
& \cdot \frac{(m+n+i)!}{m+i)!},
\end{aligned}
$$

which implies $a_{i}:=(-1)^{n+i} \frac{(m+n+i)!}{i!(n-i)!(m+i)!}$. Thus, we obtain the algebraic identity
$$
\begin{array}{l}
\sum_{i=0}^{n}(-1)^{n+i} \frac{(m+n+i)!}{i!(n-i)!(m+i)!} x(x+1) \cdots \\
\quad \cdot(x+i-1)(x+i+1) \cdots(x+n) \\
=(x-m-1) \cdots(x-m-n) .
\end{array}
$$

In particular, substituting $x=n+k$ in the above equation, and given $1 \leqslant k \leqslant m \leqslant n$, we have $m+1 \leqslant n+k \leqslant m+n$, so the right-hand side of the equation is 0. Therefore,
$$
\begin{array}{l}
\sum_{i=0}^{n}(-1)^{n+i} \frac{1}{n+k+i} \cdot \frac{(m+n+i)!}{i!(n-i)!(m+i)!} \\
\cdot k(k+1) \cdots(n+k+i-1)(n+k+i) \\
\cdot(n+k+i+1) \cdots(2 n+k)=0 .
\end{array}
$$

By canceling the factor independent of $i$,
$$
(-1)^{n} k(k+1) \cdots(2 n+k)
$$

we obtain the desired result.
Method 2: Let
$$
g(x)=\frac{(x+m+1)(x+m+2) \cdots(x+m+n)}{x+n+k} .
$$

Given $1 \leqslant k \leqslant m \leqslant n$, we have $m+1 \leqslant n+k \leqslant m+n$. Therefore, $g(x)$ is a polynomial of degree $n-1$.

By the difference formula, the $n$-th difference of $g(x)$ at 0 is
$$
\begin{array}{l}
\Delta^{n} g(0)=\sum_{i=0}^{n}(-1)^{i} \mathrm{C}_{\mathrm{m}}^{i} g(i) \\
=\sum_{i=0}^{n}(-1)^{i} \frac{n!}{i!(n-i)!} \cdot \frac{(m+n+i)!}{(m+i)!} \cdot \frac{1}{n+k+i} .
\end{array}
$$

We know that the $n$-th difference of an $(n-1)$-degree polynomial is always 0, hence
$$
\begin{array}{l}
\sum_{i=0}^{n}(-1)^{i} \frac{(m+n+i)!}{i!(n-i)!(m+i)!} \cdot \frac{1}{n+k+i} \\
=\frac{\Delta^{n} g(0)}{n!}=0 .
\end{array}
$$