Prove that if $a,b \ge2$ such that $a^b-1$ is prime, then $a=2$

Question

I want to start with a contradiction assuming $a$ isn't $2$ but no idea how to do that so I'm assuming it's wrong. Any help?

Answer 1

Hint: Use the difference of $n^{th}$ powers:

$$a^n-b^n = (a-b)\left(a^{n-1}+a^{n-2}b+a^{n-3}b^2+…+a^2b^{n-3}ab^{n-2}+b^{n-1}\right)$$

Apply this to $a^b-1$:

$$\implies a^b-1 = \color{blue}{(a-1)}\left(a^{b-1}+a^{b-2}+a^{b-3}+…+a^2+a+1\right)$$

What happens if $\color{blue}{a > 2}$?