Computing $\frac{d^k}{dx^k}\left(f(x)^k\right)$ where $k$ is a positive integer

Does anyone know a formula for the derivative $$\frac{d^k}{dx^k}\left(f(x)^k\right)$$ where $k$ is some positive integer?

I started trying to work it out but it got messy.

Apply Faà di Bruno's formula to get $$\frac{d^n}{dx^n}(g(x)^n)=\sum \frac{n!^2}{m_1!\dots m_n!(n-\sum_{j=1}^nm_j)!}g(x)^{n-\sum_{j=1}^nm_j}\prod_{j=1}^n\left(\frac{g^{(j)}(x)}{j!}\right)^{m_j},$$ where the sum is taken over the $n$-uples $(m_1,\dots,m_n)$ of integer satisfying $\sum_{k=1}^nkm_k=n$.

(I changed the notations to be conform with the link)

As a probable illustration of Davide Giraudo's powerful answer (+1) you may get the following result :

$ \begin{array} {r|cccccccc} k\\ \hline 1&1f'\\ 2&2f''f&2f'^2\\ 3&3f'''f^2&18f''ff'&6f'^3\\ 4&4f''''f^3&48f'''f^2f'&36f''^2f^2&144f'^2f''f&24f'^4\\ 5&5f'''''f^4&100f''''f^3f'&200f'''f^3f''&600 f'''f^2f'^2&900f^2f'f''^2&1200ff'^3f''&120f'^5\\ \end{array} $

This triangle appears as sequence OEIS A049009 titled "Number of functions from a set to itself such that the sizes of the pre-images of the individual elements in the range form the n-th partition in Abramowitz and Stegun order"

In fact this is a product of multinomials coefficients and 'number of multisets associated with least integer of each prime signature' (the second one shifted by $2$ I think). Perhaps that this will help a little to interpret Davide's general answer (I have not much hope for further simplification...)

Hoping it helped anyway,