**0. Background.** This question is linked to a previous one: https://math.stackexchange.com/questions/3950321/computing-sums-of-exponential-partial-bell-polynomials.
Based on the computation of the exponential partial Bell polynomial $B_{n,k}(2!,\ldots,(n-k+2)!)$ there (that I hope is correct), I managed to rewrite the problem and this led me to ask this new question.

**1. The question.** Can we explicitly compute
$$S(\alpha',\beta')=\sum_{\gamma=0}^{\min\{\beta',\,\alpha'-\beta'\}}\frac{2^{\beta'-\gamma}(-1)^\gamma}{\gamma!(\beta'-\gamma)!}\frac{(2\beta')^{\overline{\alpha'-\beta'-\gamma}}}{(\alpha'-\beta'-\gamma)!}$$
where $\beta'\geq1$ and $2\beta'\geq\alpha'\geq0$? Here $x^{\overline{k}}:=x(x+1)\ldots(x+k-1)$ denotes the rising factorial. If no closed form can be found, an estimate will be enough. Below are displayed two attempts to deal with the problem ; any advice to go further would be very appreciated.

**2.1. First attempt.** Introduce the signed Lah number (see https://en.wikipedia.org/wiki/Lah_number):
$$L(n,k):=(-1)^n\frac{n!(n-1)!}{k!(k-1)!(n-k)!}.$$
Then, putting $\gamma':=\alpha'-\beta'-\gamma$, we can write:
\begin{align*}
S(\alpha',\beta')&=\sum_{\gamma'=\min\{\alpha'-2\beta',\,0\}}^{\alpha'-\beta'}\frac{2^{2\beta'-\alpha'+\gamma'}(-1)^{\alpha'-\beta'-\gamma'}}{(\alpha'-\beta'-\gamma')!(2\beta'-\alpha'+\gamma')!}\frac{(2\beta')^{\overline{\gamma'}}}{\gamma'!}\\
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&=2^{2\beta'-\alpha'}\sum_{\gamma'=\min\{\alpha'-2\beta',\,0\}}^{\alpha'-\beta'}\frac{2^{\gamma'}(-1)^{\gamma'}}{(\alpha'-\beta'-\gamma')!(2\beta'-\alpha'+\gamma')!}\frac{(2\beta')^{\overline{\gamma'}}}{\gamma'!}\\
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&=\frac{2^{2\beta'-\alpha'}(-1)^{\alpha'-\beta'}}{(\alpha'-\beta')!(\alpha'-\beta'-1)!}\\&\quad\times\sum_{\gamma'=\min\{\alpha'-2\beta',\,0\}}^{\alpha'-\beta'}\frac{2^{\gamma'}(\gamma'-1)!}{(2\beta'-\alpha'+\gamma')!}(-1)^{\gamma'}\underbrace{(-1)^{\alpha'-\beta'}\frac{(\alpha'-\beta')!(\alpha'-\beta'-1)!}{\gamma'!(\gamma'-1)!(\alpha'-\beta'-\gamma')!}}_{=L(\alpha'-\beta',\gamma')}(2\beta')^{\overline{\gamma'}}\\
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&=\frac{2^{2\beta'-\alpha'}(-1)^{\alpha'-\beta'}}{(\alpha'-\beta')!(\alpha'-\beta'-1)!}\sum_{\gamma'=\min\{\alpha'-2\beta',\,0\}}^{\alpha'-\beta'}\frac{2^{\gamma'}(\gamma'-1)!}{(2\beta'-\alpha'+\gamma')!}(-1)^{\gamma'}L(\alpha'-\beta',\gamma')(2\beta')^{\overline{\gamma'}}
\end{align*}
The term $\frac{2^{\gamma'}(\gamma'-1)!}{(2\beta'-\alpha'+\gamma')!}$ and the lower bound $\min\{\alpha'-2\beta',\,0\}$ for the sum are annoying ; here we have something pretty close to
$$\sum_{\gamma'=0}^{\alpha'-\beta'}(-1)^{\gamma'}L(\alpha'-\beta',\gamma')(2\beta')^{\overline{\gamma'}}=(2\beta')^{\underline{\alpha'-\beta'}}.$$
Here $x^{\underline{k}}:=x(x-1)\ldots(x-k+1)$ denotes the falling factorial.

**2.2. Second attempt.** We could as well write:
\begin{align*}
S(\alpha',\beta')&=\sum_{\gamma=0}^{\min\{\beta',\,\alpha'-\beta'\}}\frac{2^{\beta'-\gamma}(-1)^\gamma}{\gamma!(\beta'-\gamma)!}\frac{2\beta'(2\beta'+1)\ldots(\alpha'+\beta'-\gamma)}{(\alpha'-\beta'-\gamma)!}\\
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&=\frac{2}{(\beta'-1)!}\sum_{\gamma=0}^{\min\{\beta',\,\alpha'-\beta'\}}2^{\beta'-\gamma}(-1)^\gamma\binom{\beta'}{\gamma}\binom{\alpha'+\beta'-\gamma}{\alpha'-\beta'-\gamma}.
\end{align*}
Now this looks like a Vandermonde's identity: https://en.wikipedia.org/wiki/Vandermonde%27s_identity. This time the annoying term is $2^{\beta'-\gamma}(-1)^\gamma$.

**EDIT.** After having read some comments/answers, I think I can handle my problem **IF** I can get a bound such as $B_{n,k}(2!,3!,\ldots,(n-k+2)!)\leq n!$ or $-$ even better $-$ $B_{n,k}(2!,3!,\ldots,(n-k+2)!)\leq(n-k+1)!$, where I computed:
<hr/>
\begin{align*}
B_{n,k}(2!,\ldots,(n-k+2)!)&=\sum_{j=0}^{\min\{k,\,n-k\}}\binom{k}{j}2^{k-j}(-1)^j\frac{(2k)^{(n-k-j)}n!}{(n-k-j)!k!}.
\end{align*}
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