This identity can be proved using the counting method. Let's consider the following problem: Select $m$ people from $n$ candidates and then ask each of these $m$ people to complete one of two tasks. How many different results are there?

One approach to compute is to first select $m$ people which has $\binom{n}{m}$ ways. Then everyone of these $m$ people has two choices, therefore there will be totally $2^m$ possibilities. Hence, the final result is $$2^m\binom{n}{m}$$

On the other hand, this problem can also be solved using casework based on the number of people who choose to complete the first of the two tasks. When there are $k$ people chose the first task, we can first select $k$ people from $n$ candidates, there are $\binom{n}{k}$ ways. Then we need to choose $(m-k)$ people from the remaining $(n-k)$ candidates who will perform the second task. Hence, there are $\binom{n}{k}\binom{n-k}{m-k}$ possibilities in total. The value of $k$ can change from $0$ to $m$. Therefore the final result is $$\displaystyle\sum_{k=0}^{m}\binom{n}{k}\binom{n-k}{m-k}$$

These two approaches should lead to the same result, i.e., $$\sum_{k=0}^{m}\binom{n}{k}\binom{n-k}{m-k}=2^m\binom{n}{m}$$