Prove that my operation is equivalent to $(n+1)!-1$

You have $$ f([a, b]) = a + b + ab = (1+a)(1+b) - 1 \\ f([a, b, c]) = a+b+c+bc+ab+ac+abc = (1+a)(1+b)(1+c) - 1 $$ and in general $$ f([x_1, x_2, \dots, x_n]) = (1+x_1)(1+x_2)\cdots(1+x_n) -1 $$ In particular $$ f([1, 2, \dots, n]) = 2 \cdot 3 \cdots (n+1) -1 = (n+1)! - 1 $$

Let

$$g(n)=f\left(\left[1,2,\dots,n\right]\right)+1$$

You can think of this as the same sum, but including the empty product.

Clearly:

$$g(0)=1=(0+1)!$$

Perhaps less obviously:

$$g(n+1)=g(n)+(n+1)g(n)$$

Since $g(n)$ contains exactly all the products that do not include $n+1$, and $(n+1)g(n)$ contain exactly all the products that do include $n+1$.

So it follows by induction that $g(n)=(n+1)!$

The different terms that you have in your sum are called the symmetrics polynoms here.

It's a classical way to link the racines of a polynom with its coefficients. Indeed in your big sum, the i-th term is a symmetric polynom that is equal to the i-th coefficient of your polynom (with a $(-1)^{n-i}$ factor).

In your case we can consider the polynom whose racines are 1,2...,n. And this polynom can also be expressed as follow:

$P(X)=(X-1)...(X-n)$,

Let's note $a_i$ the coefficient of $X^i$ in this polynom. We can now notice that

$P(-1) = (-1)^n(n+1)! = (-1)^n(1 + \sum_{i=0}^{n-1}(-1)^{n-k}a_k)$

and the term $\sum_{i=0}^{n-1}(-1)^{n-k}a_k$ is exactly your sum (convince yourself and check wikipedia page for help), and we finally find out that:

Your sum $= \sum_{i=0}^{n-1}(-1)^{n-k}a_k = (n+1)! -1$

Hope it helps.

Cheers,

Adam