For the time being, consider the function
$$f(y)=y\,\Gamma(y)\qquad \text{for} \qquad 0 \leq y \leq 1$$ and, as I did in this question of mine, let me approximate $f(y)$ by
$$g(y)=1+y(1-y) \sum_{k=0}^p d_k\, y^k$$
Using $\color{red}{p=3}$, matching the function and first derivative values at $x=0$, $x=\frac 12$ and $x=1$ , we have
$$d_0= -\gamma$$ $$ d_1=-17+6 \gamma +4 \sqrt{\pi } (\gamma +2\log (2))$$
$$d_2=4 \left(5-3 \gamma +4 \sqrt{\pi }-3 \sqrt{\pi } (\gamma +2\log (2))\right)$$
$$d_3=-4+8 \gamma -16 \sqrt{\pi }+8 \sqrt{\pi } (\gamma +2\log (2))$$
This gives
$$\int \big[f(y)-g(y)\big]^2\,dy =1.28\times 10^{-8}$$
$$\int \Big[\frac{f(y)-g(y)}y\Big]^2\,dy =3.24\times 10^{-7}$$
Now, assuming that we know the exact value of $W\left(e^{-1/e}\right)$, we have
$$\Gamma\Big(W\left(e^{-1/e}\right)\Big)\simeq1.994799$$ which seems to be quite good for a first approximation when compared to the exact $1.994832$.
If we do not know the exact value of $W\left(e^{-1/e}\right)$, we can have a reasonable approximation of it using the $[5,4]$ Padé approximant of Lambert function $W(x)$ built around $x=0$. This is
$$W(x) \sim x \frac {1+\frac{7430297 }{1597966}x+\frac{1018440443 }{156600668}x^2+\frac{1260595681
}{469802004}x^3+\frac{974868241 }{9396040080}x^4 } {1+\frac{9028263 }{1597966}x+\frac{1668309215 }{156600668}x^2+\frac{3536864687
}{469802004}x^3+\frac{14189787721 }{9396040080}x^4 }$$ which gives
$$W\left(e^{-1/e}\right)\sim 0.444019 \qquad \text{while} \qquad \text{exact}=0.444016$$
Using this approximation leads to
$$\Gamma\Big(W\left(e^{-1/e}\right)\Big)\simeq1.994784$$
Using $\color{red}{p=6}$, matching the function, first and second derivative values at $x=0$, $x=\frac 12$ and $x=1$, the above norms become $2.72\times 10^{-12}$ and $5.88\times 10^{-11}$ which is a very significant improvement.
Using the exact value of $W\left(e^{-1/e}\right)$, this gives
$$\Gamma\Big(W\left(e^{-1/e}\right)\Big)\simeq 1.99483224 \qquad \text{while} \qquad \text{exact}=1.99483209$$
Using $\color{red}{p=9}$, matching the function, first, second and third derivative values at $x=0$, $x=\frac 12$ and $x=1$, the above norms become $6.19\times 10^{-16}$ and $1.23\times 10^{-14}$ which is another significant improvement.
Using the exact value of $W\left(e^{-1/e}\right)$, this gives
$$\Gamma\Big(W\left(e^{-1/e}\right)\Big)\simeq 1.99483209100 \qquad \text{while} \qquad \text{exact}=1.99483209170$$
Remark
As in the linked question, using $p=9$ does not correspond to any overfit of function $f(y)$. A simple linear regression gives
$$\begin{array}{clclclclc}
\text{} & \text{Estimate} & \text{Standard Error} \\
d_0 & -0.577215 & 1.1 \times 10^{-8} \\
d_1 & +0.411823 & 3.7 \times 10^{-7}\\
d_2 & -0.495316 & 5.0 \times 10^{-6} \\
d_3 & +0.482941 & 3.5 \times 10^{-5} \\
d_4 & -0.477680 & 1.4 \times 10^{-4} \\
d_5 & +0.429199 & 3.6 \times 10^{-4}\\
d_6 & -0.325588 & 5.6 \times 10^{-4}\\
d_7 & +0.184385 & 5.3 \times 10^{-4} \\
d_8 & -0.066496 & 2.8 \times 10^{-4} \\
d_9 & +0.011163 & 6.1 \times 10^{-5} \\
\end{array}$$ which show the high statistical significance of the parameters.