Don't try to make yourself tired

The general expression for $n \ge 5$ is

$\displaystyle\sum_{k=5}^{n} \dfrac{k!}{(k - 5)!} = 5! \times \sum_{k=5}^{n} \dfrac{k!}{5! \times (k - 5)!} = 5! \times \sum_{k=5}^{n} \dbinom{k}{5} = 5! \times \dbinom{n + 1}{6}$ ,

where the Hockey Stick Identity was used.

In this case $n = 10$ , which leads to an answer of $5! \times \dbinom{10 + 1}{6} = 5! \times \dfrac{11!}{5! \times 6!|} = \dfrac{11!}{6!} = \boxed{55440}$ .

A simple solution -- thanks to @Spandan Senapati

$\begin{aligned} S_n & = 1\cdot 2 \cdot 3 \cdot 4 \cdot 5 + 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 + 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 + \cdots + (n-4)(n-3)(n-2)(n-1)n \\ & = \sum_{k=1}^{n-4} k(k+1)(k+2)(k+3)(k+4) \\ & = \sum_{k=1}^{n-4} \frac {{\color{#3D99F6}(k+5)}-{\color{#D61F06}(k-1)}}6 \cdot k(k+1)(k+2)(k+3)(k+4) \\ & = \frac 16 \sum_{k=1}^{n-4} \big(k(k+1)(k+2)(k+3)(k+4){\color{#3D99F6}(k+5)} - {\color{#D61F06}(k-1)}k(k+1)(k+2)(k+3)(k+4)\big) \\ & = \frac {(n-4)(n-3)(n-2)(n-1)n(n+1)}6 \end{aligned}$

For $n=10$ , we have:

$\begin{aligned} S_{10} & = \frac {(n-4)(n-3)(n-2)(n-1)n(n+1)}6 \\ & = \frac {(6)(7)(8)(9)(10)(11)}6 \\ & = \boxed{55440} \end{aligned}$