Curious combinatorial summation Let $\gamma$ denote a grid walk from the upper left corner $(1,k)$ to the lower right corner $(\ell,1)$ of the $k\times\ell$ rectangle $\{1,..,k\}\times\{1,..,\ell\}$.  There are $\binom{k+\ell-2}{k-1}$ such paths.  Denote
$$
X_\gamma = \prod_{(i,j)\in\gamma} \frac{1}{i+j-1}\,.
$$
Claim:
$$\sum_\gamma X_\gamma = \frac{1}{(k+\ell-1)(k-1)!(\ell-1)!}\,.
$$
Equivalently, and more elegantly, for a random path $\gamma$, we have: $\ \Bbb E[X_\gamma] = 1/(k+\ell-1)!$ 
Example: $k=2$, $\ell=3$. There are $3=\binom{3}{1}$ paths $\,\gamma_1:  (1,2) \to (1,1) \to (2,1) \to (3,1)$, $\,\gamma_2:  (1,2) \to (2,2) \to (2,1) \to (3,1)$, $\,\gamma_3:  (1,2) \to (2,2) \to (3,2) \to (3,1)$. Then:
$$
X_1 = \frac{1}{2\cdot 1\cdot 2\cdot 3} \ , \ X_2 = \frac{1}{2\cdot 3\cdot 2\cdot 3} \ , \ X_3 = \frac{1}{2\cdot 3\cdot 4\cdot 3} \ ,  
$$
$$X_1+X_2+X_3 = \frac{1}{12}+\frac{1}{36}+\frac{1}{72} = \frac{1}{8} = \frac{1}{4\cdot 1!\cdot 2!}\,.
$$
Question: Is there a simple proof of this combinatorial summation?  If it's known, does anyone have a reference? 
P.S.  I can in fact prove the claim but the proof is incredibly involved for such a simple looking result. 
 A: Suppose the start point is $(a,b)$ and the end point is $(c,d)$, where $a\geq c$ and $d\geq b$.
From the general form of $F((k,1),(1,l))$, and the fact that paths from $(k,1)$ go through $(k-1,1)$ or $(k,2)$, you can deduce the general form of $F((k,2),(1,l))$.
Then $F((k,3),(1,l))$ and so on.
I got this formula:
$$F((a,b),(c,d))=\frac{(a+d-b-c)!(b+c-2)!}{(a+d-1)!(a-c)!(d-b)!}$$
The base case of the induction proof is $F((a,b),(a,b))=1/(a+b-1)$ because there is one path of a single vertex.
The recursive equation is 
$$F((a,b),(c,d))=\frac1{a+b-1}\left[F((a-1,b),(c,d))+F((a,b+1),(c,d))\right]$$
$$=\frac1{a+b-1}\left[\frac{(a+d-b-c-1)!(b+c-2)!}{(a+d-2)!(a-c-1)!(d-b)!}+\frac{(a+d-b-c-1)!(b+c-1)!}{(a+d-1)!(a-c)!(d-b-1)!}\right]\\
=\frac1{a+b-1}\frac{(a+d-b-c-1)!(b+c-2)!}{(a+d-1)!(a-c)!(d-b)!}\cdot\\
\left[(a+d-1)(a-c)+(b+c-1)(d-b)\right]$$
The final factor equals $(a+b-1)(a-b-c+d)$, so the final answer is $F((a,b),(c,d))$ given above, and we only assumed $F((a,b),(c,d))$ was correct for values with a lower value of $a-b$.
