Is $x$ here undefined? $$x=Γ(0)^{(1/Γ(0))}$$ I'm aware that for the gamma function: $$\lim_{n\to^{+}0} (Γ(n)) = +\infty $$ $$\lim_{n\to^{-}0} (Γ(n)) = -\infty $$
But for $x$, the limits seem to be: $$\lim_{n\to^{-}\infty} \left(Γ(n)^{1/Γ(n)}\right) = \lim_{n\to^{+}\infty} \left(Γ(n)^{1/Γ(n)}\right) = 1 $$ I've not explored limits much, and so I'm fairly sure this is illegal.
In general, $z^w$ is defined as $\exp(w \log z)$ (where we might take different branches of the logarithm). So now the question is what kind of limit does $\Gamma(x)^{-1} \log \Gamma(x)$ have as $x \to 0$. $\Gamma(x)$ has a simple pole at $x=0$ with $\Gamma(x) \sim 1/x - \gamma$. As $x \to 0$, using any branch of the log, $\Gamma(x)^{-1} \log \Gamma(x) \sim x \log(1/x) \to 0$, so that indeed $\Gamma(x)^{1/ \Gamma(x)} \to 1$. In fact
$$\Gamma(x)^{1/\Gamma(x)} = 1 + x \ln(1/x) + O(x^2 \ln(1/x)^2)$$