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 a $K\times 1$ 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).

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).

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 a $K\times 1$ 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 $B$ 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).

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 a $K\times 1$ 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).