Q:

Summation of consecutive whole numbers with the reduction $a+b+ab$

I recently came across this problem, where you start with $n$ consecutive whole numbers; $1, 2, 3, \cdots, n$ which are arranged in any random sequence. They are then consecutively reduced* with the formula:
$$ a = a + b + ab $$
What will the the last term after reduction of all $n$ numbers is done. The answer comes out to be $(n+1)! - 1$. However, I am unable to prove the same result.
After initial reductions, I think I have to prove that
$$ \sum{n} + \sum_{1 \le i \le n}{\prod_{i \lt j \le n}{i \cdot j}} + \sum_{1 \le i \le n}{\prod_{i \lt j \le n \\ j \lt k \le n }{i \cdot j \cdot k}} + ... + n! = (n+1)! - 1 $$
but am stumped at this stage.

* Reduction is explained as follows
Remove 2 consecutive values from left ($a$ and $b$) of sequence and
  replace them with the newly computed value. After following the step
  $n - 1$ number of times, only one value will be left.

A:

We can define an operation $\bigotimes$ by $a \bigotimes b=a+b+ab=(a+1)(b+1)-1$. It is commutative and associative, so regardless of the permutation and order you choose to do reductions your reduction leaves the number $1 \bigotimes 2 \bigotimes 3 \ldots \bigotimes n=(n+1)!-1$

