Note: The double sum expression stated in OPs question is already a perfectly valid representation of the Stirling transform of the sequence $(a_n)_{n\geq 1}=\left((n-1)!\right)_{n\geq 1}$ with respect to its definition.
So, here we are looking for a different representation of the Stirling transform of $(a_n)_{n\geq 1}$ which could be regarded as more convenient or simpler according to our needs. Regrettably, a considerable simplification, e.g. reducing a sum don't presumably exist. But we establish the connection of OPs Stirling transform with the Polylogarithm $\operatorname{Li}_{1-n}(2)$ provided by Mathematica.
Let's denote the Stirling numbers of the second kind with $\begin{Bmatrix}n\\k\end{Bmatrix}$. If $(a_n)_{n\geq 1}$ is a sequence of numbers then the sequence $(b_n)_{n\geq 1}$ with \begin{align*} b_n=\sum_{k=1}^n\begin{Bmatrix}n\\k\end{Bmatrix}a_k\tag{1} \end{align*} is called the Stirling transform of $(a_n)_{n\geq 1}$.
$$$$
Stirling transform of $(a_n)_{n\geq 1}=\big((n-1)!\big)_{n\geq 1}$
According to formula (1) we obtain using an explicit expression for $\begin{Bmatrix}n\\k\end{Bmatrix}$ \begin{align*} b_n&=\sum_{k=1}^{n}\frac{1}{k!}\sum_{j=1}^k(-1)^{k-j}\binom{k}{j}j^na_k\\ &=\sum_{k=1}^{n}\frac{1}{k!}\sum_{j=1}^k(-1)^{k-j}\binom{k}{j}j^n(k-1)!\\ &=\sum_{k=1}^n\sum_{j=1}^{k}(-1)^{k-j}\binom{k-1}{j-1}j^{n-1}\tag{2} \end{align*}
$$$$
Generating functions: We introduce the generating functions $f,g$ with \begin{align*} f(x)&=\sum_{n=1}^{\infty}a_n\frac{x^n}{n!} =\sum_{n=1}^{\infty}(n-1)!\frac{x^n}{n!}=\sum_{n=1}^{\infty}\frac{x^n}{n}\\ &=-\ln(1-x)\\ g(x)&=\sum_{n=1}^{\infty}b_n\frac{x^n}{n!} \end{align*}
The representation of (2) in terms of generating functions is according to this Wiki page \begin{align*} g(x)=f(e^x-1) \end{align*} We conclude \begin{align*} g(x)&=f(e^x-1)\\ &=-\ln (2-e^x)\\ &=-\ln 2 -\ln\left(1-\frac{e^x}{2}\right)\tag{3}\\ \end{align*} Expanding the logarithmic series we obtain \begin{align*} -\ln\left(1-\frac{e^x}{2}\right)&=\sum_{j=1}^{\infty}\frac{1}{2^j}\frac{e^{jx}}{j}\\ &=\sum_{j=1}^{\infty}\frac{1}{2^jj}\sum_{m=0}^{\infty}\frac{(jx)^m}{m!}\\ &=\sum_{m=0}^{\infty}\left(\sum_{j=1}^{\infty}\frac{1}{2^j}j^{m-1}\right)\frac{x^m}{m!}\\ &=\ln(2)+\sum_{m=1}^{\infty}\left(\sum_{j=1}^{\infty}\frac{1}{2^j}j^{m-1}\right)\frac{x^m}{m!}\tag{4}\\ \end{align*}
Combining (3) and (4) we get
\begin{align*} g(x)=\sum_{m=1}^{\infty}\left(\sum_{j=1}^{\infty}\frac{1}{2^j}j^{m-1}\right)\frac{x^m}{m!} \end{align*}
We use the coefficient of operator $[x^m]$ to denote the coefficient of $x^{m}$ in $g(x)$ and obtain
\begin{align*} b_n&=\frac{1}{n!}[x^n]g(x)=\sum_{j=1}^{\infty}\frac{1}{2^j}j^{n-1}\qquad\qquad n\geq 1\tag{5} \end{align*}
$$$$
Polylogarithms: According to the definition of the Polylogarithm
\begin{align*} \operatorname{Li}_s(x):=\sum_{j=1}^{\infty}\frac{x^j}{j^s}\qquad\qquad |x|< 1, s\in\mathbb{C} \end{align*}
we observe the RHS of (5) can be written letting $x=\frac{1}{2}$ and $s=1-n$
\begin{align*} b_n=\sum_{j=1}^{\infty}\frac{1}{2^j}j^{n-1}=\operatorname{Li}_{1-n}\left(\frac{1}{2}\right) \end{align*}
Since the following identity is valid for $n\geq 1$
\begin{align*} \operatorname{Li}_{-n}(x)+(-1)^n\operatorname{Li}_{-n}\left(\frac{1}{x}\right)=0 \end{align*}
we conclude
\begin{align*} b_n=\operatorname{Li}_{1-n}\left(\frac{1}{2}\right)=(-1)^n\operatorname{Li}_{1-n}(2)\qquad\qquad n\geq 1 \end{align*}
which corresponds with the result of Mathematica.
