Solution 1: The Counting Method

Let's consider the number of shortest paths from point $A$ to $B$ in the following $(m + n -r)\times r$ grid.

 

 

On one hand, applying the basic pattern gives the answer as $$\binom{m+(n-r) + r}{r} =\binom{m+n}{r}$$

On the other hand, let's consider some mid-way points, $p_0$, $p_1$, $\cdots$, $p_r$ as shown. (These points are on the line $x+y=m$.) Any path from $A$ to $B$ must pass one of these points.

 

There are $\binom{m}{0}$ ways from $A$ to $p_0$ and $\binom{(n-r)+r}{r} = \binom{n}{r}$ ways from $p_0$ to $B$. Therefore, there are totally $\binom{m}{0}\binom{n}{r}$ ways from $A$ to $B$ via $p_0$. Similarly, there are $\binom{m}{1}\binom{n}{r-1}$ ways via $p_1$, and so on. Hence, the total number of ways from $A$ to $B$ must be $$\binom{m}{0}\binom{n}{r} + \binom{m}{1}\binom{n}{r-1} \cdots + \binom{m}{r}\binom{n}{0}$$

Therefore, we conclude the Vandermonde identity must hold.

Solution 2: The Coefficient Method

One of them is to observe the coefficient of $x^r$ in the following polynomial identity: $$(1+x)^m(1+x)^n=(1+x)^{m+n}$$

On the right side, it is clearly $C_{m+n}^r$ by the binomial theorem. The left side can be expanded to $$(C_m^0+C_m^1x+C_m^2x^2+\cdots +C_m^{m-1}x^{m-1}+C_m^mx^m)(C_n^0+C_n^1x+C_n^2x^2+\cdots +C_n^{n-1}x^{n-1}+C_n^nx^n)$$

Then for every term in the first bracket, say $x^k$, a matching term of $x^{r-k}$ must be taken from the second bracket to create an $x^r$ term with an coefficient of $C_m^kC_n^{r-k}$. Therefore, after consolidation, the coefficient of the $x^r$ term must be the left side of the given Vandermonde's identity.