$\rm [I]$ deduces $\mathbf E\left[S_n^p\right]$ based on a combinatorial reasoning.
Namely,
Result $1:$ For $p\in\mathbb N, $ with iid $X_1, X_2, \ldots, X_n,$ the $p$–th moment of sum $S_n:=\sum_{i=1}^n X_i$ is given by $$\mathbf E\left[S_n^p\right] = \sum_{q_m\in\mathcal Q^p}a_mq_m,\tag 1\label 1$$ where \begin{align}\mathcal Q^p&:=\left\{\underbrace{\mu_{p_1}\mu_{p_2}\cdots\mu_{p_m}}_{:=q_m}\mid p_1, p_2,\ldots, p_m\in \mathbb N_p~\wedge~\sum_{i=1}^m p_i= p\right\},\\a_m&:=\frac{1}{l_1!\cdots l_h!}\frac{n!}{(n-m)!}\frac{p!}{\prod_{i=1}^m p_i!} ,\end{align}
where $h$ is the number of distinct $p_i$s; $l_j$ is the number of times the $j$–th constant appears.
$\rm[II]$ also builds upon multinomial theorem a similar formula, that is equivalent to $\eqref 1.$
Result $2:$ $$\mathbf E\left[S_n^p\right] =\sum_{1\leq r\leq p} A_{r,p}{n\choose r},\tag 2\label 2$$ where $$A_{r,p}:= \sum_{\boldsymbol k\in I_{p, r}}{p\choose {k_1\ldots k_r}}{r\choose {n_1\ldots n_a}}\prod_{i=1}^a \mu_{\kappa_i}^{n_i},$$
where $I_{p,r}:= \{\boldsymbol k\in\mathbb N^r\mid k_1+\cdots k_r = p~\wedge~k_1\geq\cdots\geq k_r \}$ is the set of unordered partitioned of $p.$
Not completed
References:
$\rm [I]$ Moments of Sums of Independent and Identically Distributed Random Variables, Daniel M. Packwood, $2012,$ url: http://arxiv.org/abs/1105.6283.
$\rm [II]$ Moment equalities for sums of random variables via integer partitions and Fa`a di Bruno’s formula, Dietmar Ferger, Turkish Journal of Mathematics: Vol. $38:$ No. $3,$ Article $15,~2014,$ url: https://doi.org/10.3906/mat-1301-6.