Think about question 1 Given $k, m, n \in \mathbf{N}^{\cdot}, 1 \leqslant k \leqslant m \leqslant n$, find $\sum_{i=0}^{n}(-1)^{n+i} \frac{1}{n+k+i} \cdot \frac{(m+n+i)!}{i!(m-i)!(m+i)!}$.
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 can be written as
$$
\Delta^{n} g(0)=\sum_{i=0}^{n}(-1)^{i} C_{n}^{i} g(i)=\sum_{i=0}^{n}(-1)^{i} \frac{n!}{i!(n-i)!} \cdot \frac{(m+n+i)!}{(m+i)!} \cdot \frac{1}{n+k+i} .
$$

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

Using polynomials, we can also obtain the above result, i.e., construct the polynomial
$$
f(x)=\sum_{i=0}^{n} a_{i} x(x+1) \cdots(x+i-1)(x+i+1) \cdots(x+n)-(x-m-1) \cdots(x-m-n) .
$$

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

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

In particular, taking $x=n+k$ 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 have
$$
\begin{array}{l}
\quad \sum_{i=0}^{n}(-1)^{n+i} \frac{1}{n+k+i} \cdot \frac{(m+n+i)!}{i!(n-i)!(m+i)!} \cdot k(k+1) \cdots(n+k+i-1)(n+k+i) \cdot(n+k \\
+i+1) \cdots(2 n+k)=0 .
\end{array}
$$

By canceling the factor $(-1)^{n} k(k+1) \cdots(2 n+k)$ that is independent of $i$, we obtain the desired result.