Concrete Mathematics 6.1 Stirling Number Identities

$$\left\{\begin{array}{c}n+1\\ m+1\end{array}\right\}=\sum _k\left(\genfrac{}{}{0ex}{}{n}{k}\right)\left\{\begin{array}{c}k\\ m\end{array}\right\}\text{\hspace{1em}(6.15)}$$

We enumerate the size of the subset which $n+1$ belongs to, which equals $n-k+1$. We first choose $n-k$ elements that is in the same subset with $n+1$, which is $\left(\genfrac{}{}{0ex}{}{n}{k}\right)$, then allocate the rest $k$ elements into $m$ non-empty subsets.

$$\left[\begin{array}{c}n+1\\ m+1\end{array}\right]=\sum _k\left[\begin{array}{c}n\\ k\end{array}\right]\left(\genfrac{}{}{0ex}{}{k}{m}\right)\text{\hspace{1em}(6.16)}$$

This looks almost the same as the previous one, but it’s much harder to consider the combinatorial explanation.
What the formula do is obvious. First divide the first $n$ elements into $k$ cycles, then choose $m$ cycles from it, and combine the rest $k-m$ cycles with $n+1$ to form a new cycle. But the question is how to proof this approach is not heavy and not leaky.
Consider that we first divide the elements into $m+1$ subsets $A_1,A_2,\cdots ,A_{m+1}$. We assume that $n+1\in A_1$. Say $C(A_i)$ is a cycle formed with elements in $A_i$. Take any specific $C(A_2),C(A_3),\cdots ,C(A_{m+1})$,there are $(|A_1|-1)!$ ways to choose $C(A_1)$. In our formula, with $C(A_2),C(A_3),\cdots ,C(A_{m+1})$ fixed(after we times $k$ choose $m$), the subset $A_1$ is constitute of some of the cycles of its elements, which sum up to exactly ${\sum }_k\left[\begin{array}{c}|A_1|-1\\ k\end{array}\right]$ times, which equals $(|A_1|-1)!$.