Relationship between Fibonacci's secuence and $x^2 - x - 1$. On the end of Apostol's Mathematical Analysis' first chapter, one can find the following exercise (and I paraphrase):

Prove that the $n$-th term of the Fibonacci sequence is given by $$x_n = \frac{a^n - b^n}{a-b}$$ Where $a$ and $b$ are the roots of $x^2 - x - 1$.

Apostol also states that $x_{n+1} = x_n + x_{n-1}$. I've tried a proof by induction on $n$ in this formula. I assumed it hold, and I went on with the second step: I assume it holds for $k\leq n$ and tried to show it must hold for $n+1$:
\begin{array}{rcl}
x_{n+1} & = & x_n + x_{n-1} \\[0.2cm]
 & = & \displaystyle \frac{a^n - b^n}{a-b} + \frac{a^{n-1} - b^{n-1}}{a-b} \\
 & = & \displaystyle \frac{(a^n + a^{n-1})-(b^n + b^{n-1})}{a-b}
\end{array}
So, if I prove that $a^n + a^{n-1} = a^{n+1}$ (and the same with $b$), I should be done. So, as a lemma, I try to demonstrate this by induction, but it doesn't seem to hold.
My questions, then, are: Is there another way of proving this statement?, Is my lemma true?

Here's a large hint:
Consider the following limit:
$$\lim_{n \to \infty} \frac{F_{n+1}}{F_n}$$
Suppose that this limit exists, let's call it '$L$'.
Then
$$L = \lim_{n \to \infty} \frac{F_{n+1}}{F_n} = \lim_{n \to \infty} 1 + \frac{F_{n-1}}{F_n}$$
It can be shown that $$\lim_{n \to \infty} \frac{F_{n-1}}{F_n} = \lim_{n \to \infty} \frac{F_{n}}{F_{n+1}} = \frac{1}{L}$$
So $$L = \lim_{n \to \infty} \frac{F_{n+1}}{F_n} = 1 + \frac{1}{L}$$.
Solving for L, you will find it must satisfy:
$$L^2 - L - 1 = 0$$
