Trivial for starting $$\Gamma(x)=y\implies x=\Gamma^{(-1)}(y)$$ Now, have a look here and admire the simplicity and beauty of David W. Cantrell's solution. It is really impressive.
Defining
$$L(y)= \log \left(\frac{c+y}{\sqrt{2 \pi }}\right)$$$$\Gamma^{(-1)}(y)=\frac{L(y)}{W\left(\frac{L(y)}{e}\right)}+\frac{1}{2}$$ (using $c=0.036534$) or more rigorously $$c=\frac {\sqrt{2\pi}} e- k$$$k$ being the positive zero of the digamma function $(\approx 1.461632)$.
Trivial for starting $$\Gamma(x)=y\implies x=\Gamma^{(-1)}(y)$$ Now, have a look here and admire the simplicity and beauty of David W. Cantrell's solution. It is really impressive.
Defining
$$L(y)= \log \left(\frac{c+y}{\sqrt{2 \pi }}\right)$$ $$\Gamma^{(-1)}(y)=\frac{L(y)}{W\left(\frac{L(y)}{e}\right)}+\frac{1}{2}$$ (using $c=0.036534$) or more rigorously $$c=\frac {\sqrt{2\pi}} e- k$$ $k$ being the positive zero of the digamma function $(\approx 1.461632)$.