Kth derivative of the nth iterate of a function evaluated at a fixed point $a$?

Question

Something that has interested me is this wikipedia article: https://en.wikipedia.org/wiki/Iterated_function. Now, In the section "Some formulas for fractional iteration", a technique is provided in which you evaluate a taylor series of the function around a fixed point to find what the nth iterate is. However, this requires that you evaluate the kth order derivative of the nth iteration at a fixed point. I'm not sure how the wikipedia article went about this, which is what I'm confused about and is what this question is asking. I can understand the first order derivative, which is just a product: $f'(f^{n-1}(x))f'(f^{n-2}(x))*...*f'(x)$ and when you substitute in the fixed point, you end up with $f'(a)^{n}$. For the second order derivative, I had to define the first derivative of the nth iterate as $\prod_{i=0}^{n} f'(f^{n-i}(x))$ and use the natural log to turn the product into a sum and then took the derivative of both sides and solved for $\frac{d^2}{dx^2}f^{n}(x)$, substituted for the fixed point, and got what the wikipedia article had as the second derivative of the function. But what about further order derivatives? The wikipedia article mentions "Geometric Proggession" and I have no idea what that has to do with anything, and when attempting to find further order derivatives with the method I used to find the second derivative, things got very, very messy. I'm not sure if there is some general formula or something that can help me find $\frac{d^k}{dx^k}f^{n}(x)|_{x=a}$ where $a$ is a fixed point of $f$ or something. If such a general "formula" exists, what is it? And with that said, are there any other methods of finding the nth iterate of a function other than what was shown in the article provided?