Let $m \leq n$ be natural numbers. Starting with the product $t=m\cdot (m+1) \cdot (m+2) \cdot \cdots \cdot n$, let $T_{m, n}$ be the sum of products that can be obtained from deleting from $t$ pairs of consecutive integers (this includes $t$ itself). In the case where all the numbers are deleted, we assume the number $1$. For example, $T_{2, 7} = 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 + 2 \cdot 3 \cdot 4 \cdot 5 + 2 \cdot 3 \cdot 4 \cdot 7 + 2 \cdot 3 \cdot 6 \cdot 7 + 2 \cdot 5 \cdot 6 \cdot 7 + 4 \cdot 5 \cdot 6 \cdot 7 + 2 \cdot 3 + 2 \cdot 5 + 2 \cdot 7 + 4 \cdot 7 + 6 \cdot 7 + 1 = 5040 + 120 + 168 + 252 + 420 + 840 + 6 + 10 + 14 + 20 + 28 + 42 + 1 = 6961$. Taking $T_{n+1, n} = 1$. Show that $T_{m, n+1}=T_{m, k-1} \cdot T_{k+2, n+1} + T_{m, k} \cdot T_{k+1, n+1}$ for all $1 \leq m \leq k \leq n$.