Problem 1.1.3 in Burton's Elementary Number Theory (6th ed.) is stated as follows:
Use the Second Principle of Finite Induction to establish that for all $n \ge 1$, \begin{align} a^{n} - 1 = (a-1)(a^{n-1}+a^{n-2}+a^{n-3} + \cdots + a + 1) \end{align}
[Hint: $a^{n+1} - 1 = (a+1)(a^{n}-1) - a(a^{n-1}-1)$.]
I already came across a much simpler proof in terms of requiring fewer algebraic manipulations but it didn't use the hint given in the problem statement. Right below is my attempt at proving the statement and I wish someone could confirm what I've done.
Proof by induction.
Base case. When $n =1$, LHS = $a^{1} - 1 = a-1$ and RHS = $(a-1)(a^{1-1}) = a-1$. Thus LHS = RHS.
Inductive step. Let's assume that $ a^{k} - 1 = (a-1)(a^{k-1}+a^{k-2}+a^{k-3} + \cdots + a + 1) $ holds for $ 1, 2, \ldots k $. Then
\begin{align*} a^{k+1} - 1 &= (a+1)(a^{k}-1) - a(a^{k-1}-1) \\ &= (a+1)(a^{k}-1) - (a^{k}-a) \\ &= (a+1)(a^{k}-1) - [(a^{k} -1) + (1 -a)] \\ &= (a+1)(a^{k}-1) - (a^{k} -1) - (1 -a) \\ &= [(a+1)(a^{k}-1) - (a^{k} -1)] + (a-1) \\ &= (a^{k}-1)[a+1-1] + (a-1) \\ &= (a^{k}-1)\cdot a + (a-1) \\ &= (a-1)(a^{k-1}+a^{k-2}+a^{k-3} + \cdots + a + 1)\cdot a + (a-1) \\ &= (a-1)[(a^{k-1}+a^{k-2}+a^{k-3} + \cdots + a + 1)\cdot a + 1] \\ &= (a-1)(a^{k}+a^{k-1}+a^{k-2} + \cdots + a^{2} + a + 1) \\ \end{align*}
which is precisely the right side of the formula in the problem statement when $n = (k+1)$. Therefore by the principle of mathematical induction the formula holds for all $n \ge 1$.
The only problem with your proof is that you failed to check and confirm that for the base case, when $n=1$, the given proposition holds. Add that to your proof, and you're good to go.
It may seem utterly obvious and not worth mentioning in a proof, but without proving the base case $n=1$, your proof is meaningless in terms of providing an inductive proof.
Note also that you are indeed using the hint when you write for first line in determining $a^{k+1}$.