Initially, we are given the sequence $1,2, ... ,100$. Every minute, we erase any two numbers $u$ and $v$ and replace them with the value $uv + u + v$. Clearly, we will be left with just one number after $99$ minutes. Does this number depend on the choices that we made?
I think that the last number will be $$5050 + \sum_{i}u_iv_i$$ where $u_i$ and $v_i$ are the numbers picked at the $i$'th minute, but I'm not sure how to prove this.
It would seem that at each step the sum will be increased by the product $uv$. Is there an invariant here?
Before the first step the sum will be $$1+ \dots+ 100$$ after removing $u$ and $v$ it will be $$1+ \dots +\hat{u}+ \dots+ \hat{v}+ \dots +100 + uv + u +v = 1+ \dots +100 +uv$$ where the $\hat{u}$ means that we omit $u$. So after $99$ steps the sum will be $$1+ \dots + 100 + \sum_{i=1}^{99}u_iv_i = 5050 + \sum_{i}^{99}u_iv_i.$$
$uv + u + v = uv + u + v + 1 - 1 = (u+1)(v+1) - 1$.
For the numbers $(u+1)(v+1) - 1$ and $w$; we get $ ((u+1)(v+1) - 1)w + ((u+1)(v+1) - 1) + w = (u+1)(v+1)(w+1) - 1 $. If we continue with this idea, our result is
$$(a_1 + 1)(a_2 + 1)\cdots (a_{100} + 1) - 1 = 101! - 1.$$
Here, $\{ a_1 , a_2 , \dots , a_{100} \} = \{ 1, 2, \dots , 100 \}$. The result does not depend on the order of selection of the numbers.