By repeatedly differentiating $\Gamma(x)$, I noticed that
$$\frac{d^{n}}{{dx}^{n}}\Gamma(x)=\sum_{k=0}^{n-1}\binom{n-1}{k}\psi^{(n-k-1)}(x)\,\frac{d^{k}}{{dx}^{k}}\Gamma(x),$$
where $\psi^{(a)}(x)$ is the polygamma function. I don't find this expression useful as the derivative of $\Gamma(x)$ appears on both sides. Is it possible to establish a closed form for this derivative? Thank you.
No. Hölder (1887) showed that the field
$$K(\Gamma(s),\Gamma'(s),\ldots,\Gamma^{(n-1)}(s))$$ has transcendence degree $n$ over the field of rational functions $K=\mathbb{C}(s)$. That is, $\Gamma$ and its derivatives cannot satisfy an algebraic differential equation—but just about anything considered a "closed form" does.
That means you're stuck with the conventional form
$$\Gamma^{(n)}(s)=\Gamma(s)B_{n}(\psi(s),\psi'(s),\ldots,\psi^{(n-1)}(s))$$
where $B_n$ is the $n$th complete Bell polynomial.
Hölder, Otto. "Über die Eigenschaft der $Γ$-Funktion, keiner algebraischen Differentialgleichung zu genügen." Math. Ann. 28 (1887), 1-13.
See also
The Gamma-function is defined as
$$\Gamma(x)=\int_0^\infty t^{x-1}e^{-t} dt$$
Taking derivative with respect to $x$ yields
$$ \begin{aligned} \frac{d^n}{dx^n}\Gamma(x) &=\frac{d^n}{dx^n}\int_0^\infty t^{x-1}e^{-t} dt\\ &=\int_0^\infty \frac{d^n}{dx^n} t^{x-1}e^{-t} dt\\ &=\int_0^\infty \ln(t)^n t^{x-1}e^{-t} dt\\ \end{aligned} $$
Alternatively, based on your formula, you find
$$\Gamma^{-1}\frac{d^n}{{dx}^n}\Gamma=B_n(\psi_0,\dots,\psi_{n-1})$$
for the $n$-th Bell polynomial as K B Dave pointed out.
From your formula, you can determine recursively
$$ \begin{aligned} \Gamma^{-1}\Gamma^{(1)} &=\psi_0\\ \Gamma^{-1}\Gamma^{(2)} &=\psi_0^2+\psi_1\\ \Gamma^{-1}\Gamma^{(3)} &=\psi_0^3 + 3\psi_0\psi_1 + \psi_2\\ \Gamma^{-1}\Gamma^{(4)} &=\psi_0^4 + 6\psi_0^2\psi_1 + 3\psi_1^2 + 4\psi_0\psi_2 + \psi_3\\ \Gamma^{-1}\Gamma^{(5)} &=\psi_0^5 + 10\psi_0^3\psi_1 + 15\psi_0\psi_1^2 + 10(\psi_0^2 + \psi_1)\psi_2 + 5\psi_0\psi_3 + \psi_4\\ \Gamma^{-1}\Gamma^{(6)} &=\psi_0^6 + 15\psi_0^4\psi_1 + 45\psi_0^2\psi_1^2 + 15\psi_1^3 + 20(\psi_0^3 + 3\psi_0\psi_1)\psi_2 \\ &\quad + 10\psi_2^2 + 15(\psi_0^2 + \psi_1)\psi_3 + 6\psi_0\psi_4 + \psi_5\\ \Gamma^{-1}\Gamma^{(7)} &=\psi_0^7 + 21\psi_0^5\psi_1 + 105\psi_0^3\psi_1^2 + 105\psi_0\psi_1^3 + 70\psi_0\psi_2^2\\ &\quad + 35(\psi_0^4 + 6\psi_0^2\psi_1 + 3\psi_1^2)\psi_2 + 35(\psi_0^3 + 3\psi_0\psi_1 + \psi_2)\psi_3 \\ &\quad + 21(\psi_0^2 + \psi_1)\psi_4 + 7\psi_0\psi_5 + \psi_6 \\ \Gamma^{-1}\Gamma^{(8)} &= \psi_0^8 + 28\psi_0^6\psi_1 + 210\psi_0^4\psi_1^2 + 420\psi_0^2\psi_1^3 + 105\psi_1^4 + 280(\psi_0^2 + \psi_1)\psi_2^2\\ &\quad + 56(\psi_0^5 + 10\psi_0^3\psi_1 + 15\psi_0\psi_1^2)\psi_2 + 70(\psi_0^4 + 6\psi_0^2\psi_1 +3\psi_1^2 + 4\psi_0\psi_2)\psi_3 \\ &\quad + 35\psi_3^2 + 56(\psi_0^3 + 3\psi_0\psi_1 + \psi_2)\psi_4 + 28(\psi_0^2 + \psi_1)\psi_5 + 8\psi_0\psi_6 + \psi_7 \\ \end{aligned} $$
which aligns with the examples of the Bell polynomial.
