First Solution. (By Tiankai Liu) Let
$$f(k) = (n+1-k)!(n+k)!$$
for integers $0 \le k \le n + 1$. Note that
$$\begin{aligned} f(k) + f(k + 1) &= (n+1-k)!(n+k)! + (n-k)!(n+k+1)! \\ &= (n + 1 - k + n + k + 1)(n-k)!(n+k)! \\ &= 2(n + 1)(n-k)!(n+k)!. \end{aligned}$$
Therefore,
$$\begin{aligned} \sum_{k=0}^{n} (-1)^k (n-k)!(n+k)! &= \frac{1}{2(n+1)} \sum_{k=0}^{n} (-1)^k [f(k) + f(k+1)] \\ &= \frac{f(0) + (-1)^n f(n+1)}{2(n+1)} \\ &= \frac{(n+1)!n! + (-1)^n 0!(2n+1)!}{2(n+1)} \\ &= \frac{(n!)^2}{2} + \frac{(-1)^n (2n+1)!}{2(n+1)}. \end{aligned}$$

Second Solution. (by Gabriel Carroll) Let
$$f(x) = \sum_{k=0}^{\infty} k!x^k \quad \text{and} \quad g(x) = f(x)f(-x).$$
Note that
$$\begin{align*} f'(x) &= \sum_{k=0}^{\infty} (k+1)!(k+1)x^k \\ &= \sum_{k=0}^{\infty} (k+2)!x^k - \sum_{k=0}^{\infty} (k+1)!x^k \\ &= \frac{f(x) - x - 1}{x^2} - \frac{f(x) - 1}{x} \\ &= \frac{f(x)(1-x) - 1}{x^2} \end{align*}$$
and
$$\begin{align*} g'(x) &= f(-x)f'(x) - f(x)f'(-x) \\ &= \frac{f(-x)f(x)(1-x) - f(-x)}{x^2} - \frac{f(x)f(-x)(1+x) - f(x)}{x^2} \\ &= \frac{f(x) - f(-x)}{x^2} - \frac{2g(x)}{x}. \end{align*}$$
Denote by $[x^k]h(x)$ the coefficient of $x^k$ in a power series $h(x)$. We have
$$\begin{align*} [x^{2n}]g(x) &= \frac{1}{2n}[x^{2n-1}]g'(x) \\ &= \frac{1}{2n}[x^{2n-1}]\left(\frac{f(x)-f(-x)}{x^2} - \frac{2g(x)}{x}\right) \\ &= \frac{[x^{2n+1}](f(x)-f(-x))}{2n} - \frac{[x^{2n}]g(x)}{n}. \end{align*}$$
It follows that
$$[x^{2n}]g(x) \left(1 + \frac{1}{n}\right) = \frac{2(2n+1)!}{2n},$$
or
$$[x^{2n}]g(x) = \frac{(2n+1)!}{n+1}. \qquad (1)$$
---
On the other hand,
$$\begin{aligned} g(x) &= f(x)f(-x) = \sum_{k=0}^{\infty} k!x^k \sum_{k=0}^{\infty} (-1)^k k!x^k \\ &= \sum_{n=0}^{\infty} \left( \sum_{k=0}^{n} (-1)^{n-k} k!(n-k)! \right) x^n. \end{aligned}$$
It follows that
$$\begin{aligned} [x^{2n}]g(x) &= \sum_{k=0}^{2n} (-1)^{2n-k} k!(2n-k)! = \sum_{k=0}^{2n} (-1)^k k!(2n-k)! \\ &= 2 \sum_{k=0}^{n} (-1)^k k!(2n-k)! - (-1)^n n!^2 \\ &= 2 \sum_{m=0}^{n} (-1)^{n-m} (n-m)!(n+m)! - (-1)^n n!^2. \end{aligned}$$
The last step can be seen easily by taking $m = n - k$. Thus,
$$[x^{2n}]g(x) = 2 \sum_{k=0}^{n} (-1)^{n-k} (n-k)!(n+k)! - (-1)^n n!^2,$$
or
$$(-1)^n [x^{2n}] g(x) + n!^2 = 2 \sum_{k=0}^{n} (-1)^{-k} (n-k)!(n+k)! \quad (2)$$
Combining (1) and (2) yields
$$\begin{aligned} \sum_{k=0}^{n} (-1)^k (n-k)!(n+k)! &= \frac{(-1)^n [x^{2n}] g(x) + (n!)^2}{2} \\ &= \frac{(-1)^n (2n+1)!}{2(n+1)} + \frac{(n!)^2}{2}. \end{aligned}$$