Elegant form of $\left(\sum_{n=0}^\infty a_n x^n\right)^d$ Question
Let $(a_n)_n$ be a bounded sequence of real numbers and $x \in (0,1)$, we can then consider the series:
$$
\sum_{n=0}^\infty a_n x^n.
$$
Let now $d\in \{2,3,\dots\}$ we can then consider $\left(\sum_{n=0}^\infty a_n x^n\right)^d$. I would like to write this in the following form:
$$
\left(\sum_{n=0}^\infty a_n x^n\right)^d = \sum_{n=0}^\infty b_n x^n, \qquad (1)
$$
for certain $b_n$.
Own Attempt
From the Multinomial Theorem we find that it can be written as:
$$
\sum_{\sum_n k_n = d} \binom{d}{k_1,\dots} \prod_{t=0}^\infty (a_t x^{t})^{k_t}
=
\sum_{\sum_n k_n = d} \binom{d}{k_1,\dots} \sum_{v=0}^{\infty}\sum_{\sum_t t k_t = v} \left(\prod_{t=0}^\infty (a_t)^{k_t} \right) x^{v}
$$
we can now try to move all these sums to the left, this way we can write:
$$
\left(\sum_{n=0}^\infty a_n x^n\right) = \sum_{v=0}^\infty \left( \sum_{\sum_n k_n = d} \binom{d}{k_1,\dots} \sum_{\sum_t t k_t =v} \left( \prod_{t=0}^\infty a_t^{k_t} \right) \right) x^v
$$
we have thus found the representation $(1)$ with:
$$
b_v = \sum_{\sum_n k_n = d} \binom{d}{k_1,\dots} \sum_{\sum_t t k_t=v} \left( \prod_{t=0}^\infty a_t^{k_t}\right).
$$
I was wondering if there isn't a more elegant representation of these $b_v$?
 A: It seems you are looking for the general formula of the Cauchy product of $d$ equal power series.
Consider two power series, $\sum\limits_{n=0}^{\infty} a_n x^n$ and $\sum\limits_{n=0}^{\infty} b_n x^n$, repectively convergent in $A$ and $B$. Then its Cauchy product is defined as
$$ \left( \sum_{m=0}^\infty a_m x^m \right) \left( \sum_{n=0}^{\infty} b_n x^n \right) \doteq \sum_{k=0}^{\infty} \sum_{\ell=0}^k a_{\ell}\ b_{k-\ell}\ x^k $$
and converges at least in $A \cap B$. Letting $c_k = \sum\limits_{\ell=0}^k a_{\ell}\ b_{k-\ell}$ we see that the R.H.S. is also a power series, namely
$$ \sum_{k=0}^{\infty} c_k x^k $$
Since you are interested in the case of the Cauchy product of $d$ equal power series we write
$$ \left( \sum_{n=0}^{\infty} a_n x^n \right)^d = \prod_{k=1}^d \left(\sum_{n=0}^{\infty} a_n x^n\right) $$
Now we just apply $d-1$ times the Cauchy product definition for two power series to obtain
$$ \sum_{n_1=0}^{\infty} \sum_{n_2=0}^{n_1} \cdots \sum_{n_d=0}^{n_{d-1}} a_{n_d}\ a_{n_{d-1} - n_d}\ \cdots a_{n_1 - n_2}\ x^n $$
Thus
$$ \left( \sum_{n=0}^{\infty} a_n x^n \right)^d = \sum_{n=0}^\infty b_n x^n $$
with
$$ b_n = \sum_{n_2=0}^{n} \cdots \sum_{n_d=0}^{n_{d-1}} a_{n_d}\ a_{n_{d-1} - n_d}\ \cdots a_{n_1 - n_2} $$
A: I don't know about what you consider "elegant", but are you aware that Faà di Bruno's formula can be expressed in terms of (partial) Bell polynomials? (See also Charalambides's book.)
Applied to your problem, we have
$$\left(\sum_{n\ge 0}a_n x^n\right)^d=\sum_{n\ge0}\left(\sum_{j=0}^{d}\frac{d!}{(d-j)!}a_0^{d-j}B_{n,j}(a_1,2a_2,\dots,n!a_n)\right)\frac{x^n}{n!}$$
A: If you expand the product you will obtain a sum of products of the form $a_{n_1} a_{n_2} \cdots a_{n_d} x^{n_1+n_2+\cdots +n_d}$ which will be gathered if they share the same exponent of the $x$ variable. Hence
$$b_n = \sum_{n_1 + \cdots + n_d = n} a_{n_1} a_{n_2} \cdots a_{n_d},$$
where the sum is extended to all possible tuples $(n_1, \ldots, n_d)$ of nonnegative integers satisfying their sum is $n$.
