Using the second principle of finite induction to prove $a^n -1 = (a-1)(a^{n-1} + a^{n-2} + ... + a + 1)$ for all $n \geq 1$ The hint for this problem is $a^{n+1} - 1 = (a + 1)(a^n - 1) - a(a^{n-1} - 1)$
I see that the problem is true because if you distribute the $a$ and the $-1$ the terms cancel out to equal the left side.  However, since it is telling me to use strong induction I am guessing there is more I am supposed to be doing.  On the hint I can see that it is a true statement, but I am not sure how to use that to prove the equation or how the right side of the hint relates to the right side of the problem.  Also, I do realize that in the case of the hint $n = 1$ would be the special case. 

Sometimes the easiest way to figure out an induction argument like this is to prove a particular case.  I'll take care of proving $n = 3$, assuming you've already proven $n = 1$ and $n = 2$.
By the hint, we have
$$a^{3} - 1 = (a + 1)(a^2 - 1) - a(a - 1)$$
But the cases $n = 1$ and $n = 2$ hold, so we rewrite this as 
$$a^{3} - 1 = (a + 1)(a - 1)(a + 1) - a(a - 1)$$
Now by factoring $(a - 1)$ on the right hand side, we have
$$a^3 - 1 = (a - 1)(a^2 + 2a + 1 - a) = (a - 1)(a^2 + a + 1)$$
This is less interesting than the case $n = 4$; try to work that out on your own! After that, you should be able to work out a general $n + 1$ case.
