Don't try to make yourself tired 2

Relevant wiki: Hockey Stick Identity

Simpler solution thanks to @Spandan Senapati .

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

$\begin{aligned} \implies n-4 & = 6 \\ n & = \boxed{10} \end{aligned}$

---

Previous solution

Thanks to @Brian Charlesworth's solution in Don't try to make yourself tired .

$\begin{aligned} S & = 1\cdot 2 \cdot 3 \cdot 4 \cdot 5 + 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 + 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 + \cdots + (n-4)(n-3)(n-2)(n-1)n \\ & = \sum_{k=5}^n \frac {k!}{(k-5)!} \\ & = 5! \sum_{k=5}^n \frac {k!}{5!(k-5)!} \\ & = 5! \color{#3D99F6} \sum_{k=5}^n {k \choose 5} \quad \quad \small \color{#3D99F6} \text{By hockey stick identity} \\ & = 5! \color{#3D99F6} {n+1 \choose 6} \\ & = \frac {5! (n+1)!}{6!(n-5)!} \\ & = \frac {(n-4)\color{#3D99F6}(n-3)(n-2)(n-1)n(n+1)}6 \quad \quad \small \color{#3D99F6} \text{Given that }S = (n-3)(n-2)(n-1)n(n+1) \\ & = \frac {(n-4)\color{#3D99F6}S}6 \end{aligned}$

$\begin{aligned} \implies n-4 & = 6 \\ n & = \boxed{10} \end{aligned}$