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. 

HINT $\ $ Put $\rm\ f(n) = a^n+a^{n-1}+\:\cdots\:+1\:.\ \ $  Then
$\rm\ a^{n+1}-1\ = \ (a+1)\ (a^n-1) - a\ (a^{n-1}-1)$
$\rm\phantom{\ a^{n+1}-1\ }  =\ (a+1)\ ((a-1)\ f(n-1) - a\ (a-1)\ f(n-2))\quad $ by strong induction
$\rm\phantom{\ a^{n+1}-1\ } =\ (a-1)\ ((a+1)\ f(n-1)- a\ f(n-2))\quad$ 
$\rm\phantom{\ a^{n+1}-1\ } =\ (a-1)\ (\:f(n-1) + a\ (f(n-1)-f(n-2)) $
$\rm\phantom{\ a^{n+1}-1\ } =\ \ldots $
$\rm\phantom{\ a^{n+1}-1\ } =\ (a-1)\ f(n)$
