6. Let $a \star b=a b+a+b$ for all integers $a$ and $b$. Evaluate $1 \star(2 \star(3 \star(4 \star \ldots(99 \star 100) \ldots)))$.
Answer: $101!-1$ We will first show that $\star$ is both commutative and associative.
- Commutativity: $a \star b=a b+a+b=b \star a$
- Associativity: $a \star(b \star c)=a(b c+b+c)+a+b c+b+c=a b c+a b+a c+b c+a+b+c$ and $(a \star b) \star c=(a b+a+b) c+a b+a+b+c=a b c+a b+a c+b c+a+b+c$. So $a \star(b \star c)=(a \star b) \star c$.

So we need only calculate $((\ldots(1 \star 2) \star 3) \star 4) \ldots \star 100)$. We will prove by induction that
$$
((\ldots(1 \star 2) \star 3) \star 4) \ldots \star n)=(n+1)!-1 \text {. }
$$
- Base case $(n=2):(1 \star 2)=2+1+2=5=3$ ! -1
- Inductive step:

Suppose that
$$
(((\ldots(1 \star 2) \star 3) \star 4) \ldots \star n)=(n+1)!-1 .
$$

Then,
$$
\begin{aligned}
((((\ldots(1 \star 2) \star 3) \star 4) \ldots \star n) \star(n+1)) & =((n+1)!-1) \star(n+1) \\
& =(n+1)!(n+1)-(n+1)+(n+1)!-1+(n+1) \\
& =(n+2)!-1
\end{aligned}
$$

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Hence, $((\ldots(1 \star 2) \star 3) \star 4) \ldots \star n)=(n+1)!-1$ for all $n$. For $n=100$, this results to $101!-1$.