calculus - Elegant form of $left(sum_{n=0}^infty a_n x^nright)^d$

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 $b_n$.

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 $b_v$?

Answer

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

$$\left(\sum_{n\ge 0}a_n x^n\right)^d=\sum_{n\ge0}\left(\sum_{j=0}^{d}\frac{d!}{(d-j)!}a_0^{d-j}B_{n,j}(a_1,2a_2,\dots,n!a_n)\right)\frac{x^n}{n!}$$