Convergence of fixed-point iteration for $p$ times continuously differentiable function I am stuck at this problem:

Let $\alpha\in\Bbb{R}$ be some number that satisfies $g(\alpha)=\alpha$ for some function $g$ that is  $p$ times continuously differentiable on some neighborhood of $\alpha$ for some constant $1\leq p\in\Bbb{N}$.
Suppose that $g'(\alpha) =g''(\alpha)=...=g^{(p-1)}(\alpha)=0 $ and that $g^{(p)}(\alpha)\neq 0$
Now let $x_n$ be the sequence defined by $x_n=g(x_{n-1})$ for all $1\leq n\in\Bbb{N}$ ($x_n$ is a fixed-point iterations sequence)
Prove that if $x_0$ is "close" to $\alpha$ then the order of convergence of $x_n$ is $p$.
(That is: We need to show that there exist a positive constant $\lambda$ such that $\lim_{n\to \infty}\frac{|x_{n+1}-q|}{|x_n-q|^p}=\lambda$ where $q=\lim_{n\to \infty}x_n$)
And show that $\lim_{n\to \infty}\frac{\alpha-x_{n+1}}{(\alpha-x_n)^p}=(-1)^{(p-1)}\frac{g^{(p)}(\alpha)}{p!}$
(Hint: Expand $g$ to Taylor polynomial to get $g(x)=\Sigma_{k=0}^{p-1}\frac{f^{(k)}(\alpha)}{k!}(x-\alpha)^k + \frac{f^{(p)}(\zeta(c))}{(p+1)!}(x-\alpha)^{(p+1)}$ (where $c$ is between $x$ and $\alpha$) and use the fact that $x_0$ is close to $\alpha$ to conclude that $ \frac{f^{(p)}(\zeta(c))}{(p+1)!}(x-\alpha)^{(p+1)}\approx 0$)

I've tried several ways but none led me to a solution.
Thanks for any hint/help.
 A: The solution I will give is an extension of the one I provided in this question.  However, it will take into account the higher $p$.  
We are given that $g(\alpha) = \alpha$ and that $x_{n+1} = g(x_n)$ is a sequence that converges to $\alpha$ (i.e. to the fixed point).  The limit we are interested in calculating can be viewed as the ratio of two $p$ times continuously differentiable functions: $\alpha - g(x)$ and $(\alpha - x)^p$.  We can evaluate the limit of $x_n$ then as follows:
$$\lim_{n \to \infty} \frac{\alpha - x_{n+1}}{(\alpha - x_n)^p} = \lim_{x \to \alpha} \frac{\alpha - g(x)}{(\alpha - x)^p}$$
The numerator and denominator both limit to zero, so by L'Hospital's rule:
$$\lim_{x \to \alpha} \frac{\alpha - g(x)}{(\alpha - x)^p} = \lim_{x \to \alpha} \frac{(-1)g'(x)}{(-1)p(\alpha - x)^{p-1}}$$
However, by assumption, the derivatives of $g$ up to and including $(p-1)$ are all zero at $\alpha$.  By iterating L'Hospital's rule we arrive at:
$$\lim_{x \to \alpha} \frac{\alpha - g(x)}{(\alpha - x)^p} = \lim_{x \to \alpha} \frac{(-1)g^{(p)}(x)}{(-1)^pp!} = (-1)^{(p-1)}\frac{g^{(p)}(\alpha)}{p!}$$
