Since $(a+1)(b+1) = c+1$, where $c$ is a number that would be written on the board instead of $a$ and $b$, the product of the numbers that are greater by $1$ than the numbers on the board does not change in the process. In the beginning the product is $2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7$ or $5040$. Consequently the only number that can be on the board in the end is $5040 - 1 = 5039$.

Let us define $x * y = xy + x + y$ for every two numbers $x$ and $y$, then in each step Juku substitutes some $a$ and $b$ with $a * b$. The binary operation $*$ is commutative and associative, due to which the final result is independent on the order the numbers are being deleted. Therefore in the end the board will contain one number that equals $((1 * 2) * (3 * 4)) * (5 * 6) = (5 * 19) * 41 = 119 * 41 = 5039$.