Simplify sum $\sum_{m_1,m_2,\ldots,m_k}{p \choose m_1\; m_2 \; \ldots \; m_k}\Pi_{i=1}^k (2m_i-1)!!$ for $p\to \infty$ Let $1 \le k \le p$ be integers. Define
$$
S_{p,k} := \sum_{m_1,m_2,\ldots,m_k}{p \choose m_1\; m_2 \; \ldots \; m_k}\Pi_{i=1}^k (2m_i-1)!!,
$$
where the sum if over all nonnegative $k$-partitions $(m_1,\ldots,m_k)$ of $p$, and
$n!!$ is the product of all integers between $n$ and $1$ inclusive, with the same parity as $n$, and $n!!=1$ if $n \le 0$.

Question. Does $S_{p,k}$ simplify (say via special functions) ?

Note. I'm fine with asymptotic formula for $S_{p,k}$ in the limit $p \to \infty$.
Edit: motivation of the question
Let $T_{n,d,p}:= \mathbb E[\mbox{Tr}(WW^\top)^p]=\mathbb E[(\sum_{i=1}^n\sum_{j=1}^jw_{ij}^2)^{p}]$ be the expectation of the $p$th power of the trace of the Wishart matrix $WW^\top$, shere $W$ is an $n \times p$ matrix with iid entries from $N(0, 1)$. Via a quick-and-dirty calculation, I it can be seen that $T_{n,d,p} = \sum_{k=1}^{\min(p,n)}{n \choose k}S_{p,k}$. This is because the general term in the sum $(\sum_{i=1}^n\sum_{j=1}^jw_{ij}^2)^{p}$ is $(w_{i_1,*}^2)^{m_k} \ldots  (w_{i_k,*}^2)^{m_1}$ for distinct $i_1,\ldots,i_k \in [n]$ and $k \le \min(n,p)$, and the expectation of this term is $\Pi_{i=1}^k(2m_i-1)!!$, which only depends on the numbers $m_1,\ldots,m_k$, and not on the specific $i$'s!
 A: Following Phicar's very insightful comment, it is clear that
$$
S_{p,k}=\mathbb E\Big[(Z_1^2+Z_2^2+\dots+Z_k^2)^p\Big]
$$
where $Z_1,\dots,Z_k$ are independent standard normal variables. That is, $S_{p,k}$ is the $p^{th}$ moment of a $\chi^2$ random with $k$ degrees of freedom! Once you know the moment generating function of such a random variable (see here for a derivation), you can read off the moments by expanding into a Taylor series. The result is that
$$
S_{p,k}=2^p\;\Gamma(p+\tfrac{k}2)/\Gamma(\tfrac{k}2)=(k+2p-2)(k+2p-4)\cdots(k+2)k
$$
As a check, when $p=2$, every composition is a permutation of $(2,0,\dots,0)$ or $(1,1,0,\dots,0)$. There are $k$ of the former, each contributing $\binom{2}{2,0,\dots,0}(2\cdot 2-1)!!=3$, and there are $\binom{k}2$ of the latter, each contributing $\binom{2}{1,1,0,\dots,0}1!!\cdot 1!!=2$. Therefore, $S_{3,k}=3k+2\binom{k}2$, which is indeed equal to $k(k+2)$.

Again, following Phicar's prompt, one can give a combinatorial proof as well. Both sides of the equation
$$
\sum_{m_1+\dots+m_k=p}\binom{p}{m_1,\dots,m_k}\prod_i (2m_i-1)!!=k(k+2)(k+4)\cdots(k+2(p-1))
$$
count the number of ways to take $p$ couples, assign each couple one of $k$ colors, and then partition the set of people in each color into pairs. That is, in the end, the $2p$ initial people will all be divided into pairs, and each pair will have a color, in such a way that two people who were initially in the same couple must end up with the same color.
This interpretation is a direct reading of the LHS; if you want $m_i$ people to have color number $i$ for each $i\in \{1,\dots,k\}$, then there are $\binom{p}{m_1,\dots,m_k}$ ways to assign the colors, and then $(2m_i-1)!!$ ways to form a perfect pairing within the $i^{th}$ color.
For the RHS, imagine assigning both the colors and pairings for each couple, going one couple at a time.

*

*For the first couple, there are $k$ ways to choose their color. You then temporarily pair these two together.


*For the second couple, there are $k+2$ options. For $k$ of these options, you pair off this couple with each other, and give them one of $k$ colors. For the other two options, they will trade partners with the previous couple, and take their color. This can be done in two ways; if the couples are $(A,B)$ and $(C,D)$, you can pair $C$ with either $A$ or $B$, and then $D$ gets the other person in $(A,B)$.


*For each subsequent couple, the logic is the same. They can either pair together and get a new color in $k$ ways, or they can trade partners with any of the previously formed pairs. If we are assigning couple number $i$ ($1\le i\le p$), then there are $2(i-1)$ ways to do this, as one of the new couple members can "steal" the partner of any previous pair, and then the other couple member gets the member of the previous pair.
Therefore, there are $p$ stages, and $k+2(i-1)$ ways to ways to complete the $i^{th}$ stage, so in total there are $k(k+2)(k+4)\cdots(k+2(p-1))$ ways. QED.
