I expect that this question is an elementary exercise in combinatorics, so hopefully somebody who knows more than me can explain. 


Specifically, if $m\in\mathbb{N}$ and
$$f(x)=\sum\_{n=1}^{\infty}a\_nx^n,$$
one obviously has 
$$f^m(x)=\sum\_{n=m}^{\infty}\left(\sum\_{k\_1+k\_2+\cdots + k\_m=n}a\_{k\_1}a\_{k\_2}\cdots a\_{k\_m}\right)x^n.$$


The number of distinct products appearing in the iterated convolution is the integer partition number $p(n-m)$ so, writing the inner sum as 


(1)
$$\sum\_{j\_1+2j\_2+\cdots + mj\_m=n}s\_{n}(j\_1,j\_2,...,j\_m)a^{j\_1}\_1a^{j\_2}\_2\cdots a^{j\_m}\_m,$$
the questions arise: 


What is the coefficient sequence of integers $s\_{n}(j\_1,j\_2,...,j\_m)$, explicitly? 


What is its combinatorial significance? 


EDIT: The fact that the number of terms in the sum is $p(n-m)$ is perhaps not so obvious, but you can reason as follows: note that the largest subscript that can appear in the coefficient of $x^n$ is $n-m+1$, because $n-m+1+(m-1)\times 1=n$. This restricts you to having only those partitions of $n$ whose largest part is $m-n+1$, that is, the set of solutions to the diophantine equation


(2)$$ j\_1+2j\_2+\cdots +(n-m+1)j\_{n-m+1}=n.$$
On the other hand, since there are a total of $m$ coefficients in each product (counting repetitions), you know that 


(3)$$j\_1+j\_2+\cdots +j\_{n-m+1}=m.$$
Thus, subtracting (2) from (1), you get $$j\_2+2j\_3+\cdots +(n-m)j\_{n-m+1}=n-m,$$
the number of distinct solutions to which is the partition number $p(n-m)$. I am trying to think of a better notation for the summation in (1), as I think it is definitely misleading.


EDIT: I should've indexed everything by $+1$ so that (1) becomes
$$\sum\_{j\_1+2j\_2+\cdots +(n-m)j\_{n-m}=n-m}s(j\_1,j\_2,...,j\_{n-m})a\_1^{m-(j\_1+\cdots +j\_{n-m})}a\_2^{j\_1}\cdots a\_{n-m+1}^{j\_{n-m}}.$$



What are the coefficients when you write iterated additive convolutions as sums over integer partitions?
Let $j\_1 + 2j\_2 + \cdots + nj\_n = n$. The coefficient of $a\_1^{j\_1} a\_2^{j\_2} \ldots a\_n^{j\_n} x^n$ in $ \Bigl( \sum\_{n=1}^\infty a\_n x^n \Bigr)^m $ is equal to the multinomial coefficient


$$ \binom{m}{j\_1,j\_2,\ldots,j\_n} $$


since when we multiply out $\Bigl( \sum\_{n=1}^\infty a\_n x^n \Bigr) \ldots \Bigl( \sum\_{n=1}^\infty a\_n x^n \Bigr) $ we must take $a\_ix^{i}$ from exactly $j\_i$ of the terms.
So the coefficient is
$\frac{m!}{j\_1! j\_2! \ldots j\_n!} $
provided $j\_1 + \cdots + j\_n = m$ and $0$ otherwise. 


(There are some problems with the $s$ notation in the question. I think $s\_n(j\_1,\ldots,j\_m)$ should be $s\_m(j\_1,\ldots,j\_n)$, since it is possible for $a\_rx^r$ to contribute to the coefficient of $x^n$ in $f(x)^m$ for $r > m$. Later it is noted that the highest $r$ that can appear is $n-m+1$, but the shifted definition then defines a different family of numbers.)


