Determining PDF on a given compliacted moments

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Suppose the random variable $X$ is known to have MGF and its $k$-th moment is given as \begin{equation} \label{eqn:moments} m_k = (-1)^k \sum_{l=1}^k \sum_{ \ \ \ \ \ \ \ \ j_i\ge 1 \\ j_1+\cdots +j_l = k}\binom{k}{j_1 \ j_2 \ \cdots \ j_l }\binom{-2}{l}2^l \end{equation}

Once the moments are simplified and is possible to evaluate the MGF, the PDF is derived(taking the inverse of Laplacian or whatever). However, the real problem is that there is a difficulty to simplify $m_k$. The form is very analogous to the PDF of a multivariate Bernoulli distribution, but the condition $j_i\ge 1$ let just stuck in here.

Appreciate any helps.

EDIT: I have found there is a simplified version of $m_k$(see below reference) and seems using Stirling number is the best possible way. From here, is it possible to evaluate the closed-form of MGF noting that $$ M_X(t) = \sum_{k=0}^\infty m_k\frac{t^k}{k!} \ \ \ ?$$

How to simplify $\sum _{l=1} ^{k} \sum_{j_1 \ge 1} \cdots \sum_{j_l \ge 1, j_1 + \cdots + j_l=k} \binom{k}{j_1 \cdots j_l} \binom{-2}{l} 2^l$