How to prove Vandermonde's Identity: $\sum_{k=0}^{n}\binom{R}{k}\binom{M}{n-k}=\binom{R+M}{n}$? How can we prove that

$$\sum_{k=0}^{n}\binom{R}{k}\binom{M}{n-k}=\binom{R+M}{n}?$$

(Presumptive) Source: Theoretical Exercise 8, Ch 1, A First Course in Probability, 8th ed by Sheldon Ross.

Consider the $K\times K$ matrix 
$$B= \left[ \begin{array}{ccccccc}
1 & 0 & 0 & \cdots & 0 & 0 & 0 \\
1 & 1 & 0 & \cdots & 0 & 0 & 0 \\
0 & 1 & 1 & \cdots & 0 & 0 & 0 \\
0 & 0 & 1 & \cdots & 0 & 0 & 0 \\
& & & \ddots & & & \\
0 & 0 & 0 & \cdots & 1 & 1 & 0 \\
0 & 0 & 0 & \cdots & 0 & 1 & 1 \\
\end{array}
\right]$$
When it is multiplied by $K\times 1$ vectors, which starts with k-1st row of Pascals's triangle, the result is a $K\times 1$ vector which contains the 1st $K$ elements of kth row of Pascal's triangle.  This is because it effectively mimics addition of elements of k-1st row of Pascal's triangle to produce kth row of Pascal's triangle.  Except that it only does it for the 1st $K$ elements of a row.  In particular, 
$$B \times \left[ \begin{array}{c}
1 \\
0 \\
\vdots \\
0
\end{array}
\right]
=\left[ \begin{array}{c}
1 \\
1 \\
\vdots \\
0
\end{array}
\right]
$$
$$
B\times B \times 
\left[ \begin{array}{c}
1 \\
0 \\
\vdots \\
0
\end{array}
\right]
=B\times \left[ \begin{array}{c}
1 \\
1 \\
\vdots \\
0
\end{array}
\right]
=\left[ \begin{array}{c}
1 \\
2 \\
1 \\
\vdots \\
0
\end{array}
\right]
$$
It is an easy proof that all powers of $B$ are symmetric with respect to their antidiagonal (just consider $B$ as a sum of $I$ and $N=B-I$ and then look at the expansion of $(I+N)^m$).  If we designate $\left[ \begin{array}{c}
1 \\
0 \\
\vdots \\
0
\end{array}
\right]$
as the zeroth row of Pascal's triangle, then the vector containing the 1st $K$ elements of $m$'s row of Pascal's triangle is $$B^m \times \left[ \begin{array}{c}
1 \\
0 \\
\vdots \\
0
\end{array}
\right],$$
which is the 1st column of $B^m$ (and by the symmetry, the last row of $B^m$).
Now set the $K$ (of this answer) equal to $n$ (of the posited question).
Then the left-hand side of the identity can be considered to be the dot product of $nth$ row of $B^R$ and 1st column of $B^M$, which is the $(n,1)$ element of $B^{R+M}$.  Which also happens to be the right-hand side of the identity (in the posited question).
