To choose a subset $A$ of size $4$ in $\{1,\dotsc,8\}$, we can start by choosing a number $m\in\{4,5,6,7,8\}$ which will be the maximum element of $A$. We then have to choose $3$ more elements, which must lie in $\{1,\dotsc,m-1\}$ because $m$ is the maximum. There are $\small\left(\begin{matrix}m-1\\3\end{matrix}\right)$ ways to do this, giving $\small\left(\begin{matrix}m-1\\3\end{matrix}\right)$ different sets of size $4$ in $\{1,\dotsc,8\}$ whose maximum is $m$. From this we see that $$ \left(\begin{matrix}8\\4\end{matrix}\right) = \left(\begin{matrix}3\\3\end{matrix}\right) + \left(\begin{matrix}4\\3\end{matrix}\right) + \left(\begin{matrix}5\\3\end{matrix}\right) + \left(\begin{matrix}6\\3\end{matrix}\right) + \left(\begin{matrix}7\\3\end{matrix}\right) = \sum_{m=4}^8 \left(\begin{matrix}m-1\\3\end{matrix}\right). $$ By counting subsets of size $k$ in $\{1,\dotsc,n\}$ in the same way, we see that $ \left(\begin{matrix}n\\k\end{matrix}\right) = \sum_{m=k}^n \left(\begin{matrix}m-1\\k-1\end{matrix}\right).$