The original problem can be found here, 100 prisoners problem , where $n=100$ and $n/2 \lt k \le n$. Letting $L(n,k)$ equal the number of permutation on $n$ distinct objects with greatest cycle length $k$. The formula for $L(n,k)$ is:
$$\begin{pmatrix}n\\k\end{pmatrix}\cdot(k-1)!\cdot(n-k)!=\frac{n!}{k}$$
The remaining $n-k$ elements can occur in any of $(n-k)!$ permuted orders. This is where I need some clarification as to what $(n-k)!$ means. Letting $r=n-k$, then $(n-k)!$ is just your standard permutation.
$$r\cdot(r-1)\cdot(r-2)\dots(2)\cdot(1)=r!$$
Since the remaining elements can form cycles of sizes from $1$ to $k-1$ then how do we know this equals to $r!$ ? This reference, On the Number of Permutations on $n$ Objects with Greatest Cycle length $k$, states it does by the following formula. NOTE, I reversed the notation from $L(k,n)$ in the paper to $L(n,k)$ in this post.
$$\sum_{i=1}^rL(r,i) = r!$$
Is there some nice explanation for this or am I just mixed up and over thinking this?
Second, and more important to me is the possibility of reformulating the original $L(n,k)$ formula so that you start the formula with the elements that remain instead of accounting for them at the end? I am think of something along the lines of the following and reordering seems to matter but I can't see a way forward when doing this.
$$\begin{pmatrix}n\\1\end{pmatrix}\cdot\begin{pmatrix}n-1\\1\end{pmatrix}\dots\begin{pmatrix}n-k\\1\end{pmatrix}\cdot L(_,k)+\dots+$$