Combinatorics proof on $Q_n$

Question

Let $Q_n$ denote the number of permutations of the set {1,2,3,$\dots$,n} in which none of the patterns 12,23,34,$\dots$(n-1)n occurs. It is know that $$Q_n=n!-{n-1\choose1}(n-1)!+{n-1\choose2}(n-2)!-{n-1\choose3}(n-3)!+ . . .+(-1)^{n-1}{n-1\choose{n-1}}1!$$ Poove that this can be rewritten as

$$Q_n=(n-1)!\left(n-\frac{n-1}{1!}+\frac{n-2}{2!}-\frac{n-3}{3!}+ . . . +(-1)^{n-1}\frac{1}{(n-1)!}\right)$$

Answer 1

HINT: $$\begin{align*} \binom{n-1}k(n-k)!&=\frac{(n-1)!(n-k)!}{k!(n-1-k)!}\\ &=\frac{(n-1)!(n-k)}{k!}\\ &=(n-1)!\cdot\frac{n-k}{k!} \end{align*}$$