Example 8 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 reasoning process.
(2000 China National Training Team Selection Exam Problem)
Solution 1: Construct a polynomial
$$
\begin{aligned}
f(x)= & \sum_{i=0}^{n} a_{i} x(x+1) \cdots(x+i-1)(x+i+1) \cdots(x+n)-(x-m-1)(x-m-2) . \\
& (x-m-3) \cdots(x-m-n),
\end{aligned}
$$

We can appropriately choose the coefficients \(a_{i}(0 \leqslant i \leqslant n)\) such that \(f(x) \equiv 0\).
Since \(f(x)\) is a polynomial of degree at most \(n\), \(f(x) \equiv 0\) 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
$$
\begin{aligned}
0=f(-i)= & a_{i}(-i)(-i+1) \cdots(-i+i-1)(-i+i+1) \cdots(-i+n)- \\
& (-i-m-1)(-i-m-2) \cdots(-i-m-n) \\
= & a_{i}(-1)^{i} \cdot i!(n-i)!-(-1)^{n} \frac{(m+n+i)!}{(m+i)!},
\end{aligned}
$$

which implies \(a_{i}=(-1)^{n+i} \frac{(m+n+i)!}{i!(n-i)!(m+i)!}\).
Thus, we obtain the algebraic identity
$$
\begin{aligned}
& \sum_{i=0}^{n}(-1)^{n+i} \frac{(m+n+i)!}{i!(n-i)!(m-i)!} x(x+1) \cdots(x+i-1)(x+i+1) \cdots(x+n) \\
= & (x-m-1)(x-m-2) \cdots(x-m-n) .
\end{aligned}
$$

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

Dividing both sides by \((-1)^{n}(n+k)(n+k+1) \cdots(2 n+k)\), we obtain
$$
\sum_{i=1}^{n}(-1)^{i} \frac{1}{n+k+i} \cdot \frac{(m+n+i)!}{i!(n-i)!(m+i)!}=0 .
$$