Given a polynomial $f$ of degree $m$: $$f(x)=\sum_{j=0}^{m} a_jx^j$$
Now this polynomial is applied to itself $f(f(f(f...(f(x))))$ for $i$($-1$) times $->f^i(x)$. The resulting function is a polynomial as well but has a degree of $m^i$
$$f^1(x)=f(x)$$ $$f^2(x)=f(f(x))$$ $$f^i(x)=\sum_{k=0}^{m^i} b_{i,k}x^k$$
Is there a short form for each coefficient? And with this a short form of $f^i(x)$?
Trials so far
It should look something like (other form also possible):
$$b_{i,k} = \sum_{j=0}^{m^i} \prod^?_? a_j^?$$
It starts with: (edit: also complicated)
and ends with: $$b_{i,m^i} = a_m^i$$
For other $k$ it gets complicated.
Alternative form: $$f^i(x) =\sum_{j=0}^m a_j [..\text{i times sum total}...\sum_{j=0}^m a_j\{\sum_{j=0}^m a_j(\sum_{j=0}^m a_jx^j)^j\}^j]$$