Remarks: Numerically, $\Gamma(W(3)^{-W(3)} + 1) - \Gamma(3^{-3} + 1) \approx 0.0000044051$. Thus, we need good estimates.
First, we estimate $W(3)$.
$W(3)$ is the solution of $y\mathrm{e}^y = 3$. Clearly $1< W(3) < 2$.
Fact 1: $y\mathrm{e}^y \ge \frac{2\mathrm{e}y^2 + 4\mathrm{e}y}{y^2 - 6y + 11}$ for all $y \ge 1$.
Let $g(y) = \frac{2\mathrm{e}y^2 + 4\mathrm{e}y}{y^2 - 6y + 11}$.
Then, we have $g(W(3)) \le 3$ which results in
$$W(3) \le \frac{-2\mathrm{e} - 9 + \sqrt{4\mathrm{e}^2 + 102\mathrm{e} - 18}}{2\mathrm{e} - 3} < \frac76 . $$
Second, we estimate $W(3)^{-W(3)}$.
Fact 2: $x^{-x} \ge 1 - (x - 1) + \frac37(x - 1)^3$ for all $x \in [1, 7/6]$.
(The RHS is based on the Taylor approximation of the LHS.)
Using Fact 2, we have
$$W(3)^{-W(3)} > \frac{3621}{3811}.$$
Third, since $x\mapsto \Gamma(x + 1)$ is strictly increasing on $x>1/2$, it suffices to prove that
$\Gamma(28/27) < \Gamma(7432/3811)$
or
$$\frac{1}{27}\Gamma(1/27) < \frac{3621}{3811}\Gamma(3621/3811).$$
Using $\Gamma(x)\Gamma(1 - x) = \frac{\pi}{\sin (\pi x)}$,
we have $\Gamma(3621/3811)\Gamma(190/3811) = \frac{\pi}{\sin \frac{190\pi}{3811}}$.
It suffices to prove that
$$\frac{1}{27}\Gamma(1/27)\Gamma(190/3811) < \frac{3621}{3811}\, \frac{\pi}{\sin \frac{190\pi}{3811}}$$
or
$$- \ln \Gamma(1/27) - \ln \Gamma(190/3811) + \ln \left(\frac{3621}{3811}\, \frac{27 \pi}{\sin \frac{190\pi}{3811}}\right) > 0. \tag{1}$$
Using $\ln(1 + u) - u \ge - \frac12 u^2 + \frac13 u^3 - \frac14 u^4$ for all $u \ge 0$ and
$$-\ln \Gamma(x) = \ln x + \gamma x +
\sum_{n=1}^\infty [\ln(1 + x/n) - x/n],$$
we have
\begin{align*}
-\ln \Gamma(1/27) &\ge \ln \frac{1}{27} + \frac{1}{27}\gamma
+ \sum_{n=1}^\infty \left(- \frac12\,\frac{1}{27^2n^2} + \frac13\,\frac{1}{27^3n^3} - \frac14\,\frac{1}{27^4 n^4}\right)\\
&= \ln \frac{1}{27} + \frac{1}{27}\gamma - \frac{1}{8748}\pi^2 + \frac{1}{59049}\zeta(3) - \frac{1}{191318760}\pi^4, \tag{2}
\end{align*}
and
\begin{align*}
&-\ln \Gamma(\tfrac{190}{3811})\\
\ge\, &
\ln \frac{190}{3811} + \frac{190}{3811}\gamma
+ \sum_{n=1}^\infty \left(- \frac12\,\frac{190^2}{3811^2n^2} + \frac13\,\frac{190^3}{3811^3 n^3} - \frac14\,\frac{190^4}{3811^4 n^4}\right)\\
=\,& \ln \frac{190}{3811} + \frac{190}{3811}\gamma - \frac{9025}{43571163}\pi^2 + \frac{6859000}{166049702193}\zeta(3)\\[8pt]
&\quad - \frac{32580250}{1898446245172569}\pi^4 \tag{3}
\end{align*}
where $\gamma$ is the Euler-Mascheroni constant, and $\zeta(3) = \sum_{n=1}^\infty \frac{1}{n^3}$.
Finally, using (2) and (3), the inequality (1) is verified: we get
$$- \ln \Gamma(1/27) - \ln \Gamma(190/3811) + \ln \left(\frac{3621}{3811}\, \frac{27 \pi}{\sin \frac{190\pi}{3811}}\right) > 0.0000017.$$
We are done.
