Let $$\Gamma^{(n)}(x) \quad(n \in \mathbb{N}, x \in \mathbb{R})$$ be the notation for the Gamma Function $\Gamma(x)$ used $n$ times on itself.
(Example: $\Gamma^{(3)}(x) = \Gamma(\Gamma(\Gamma(x)))$
Why is $\lim_{n \to \infty} \Gamma^{(n)}(x) = 1$ for $x \in [1,2]$?
It think it could have something to do with the fact that $0 < x! \leq 1$ for $x \in [0,1]$ or with the definition that $0! = 1! = 1$, but I can't prove it. Can someone help me?
As was mentioned by Vercassivelaunos in comments, it's sufficient to prove that $\left|\Gamma'(x)\right| \leq c < 1$ for $x\in[1, 2]$. It's known that $$ \Gamma^{(n)}(x) = \int_0^{\infty} t^{x-1} e^{-t} \left(\ln t\right)^n \mathrm{d}t, $$ so $\Gamma '' (x) > 0$ for all $x > 0$ and hence $\Gamma'(x)$ is increasing for $x > 0$. Let's find $\Gamma'(1)$ and $\Gamma'(2)$. From the properties of the digamma function $\psi(x) = \frac{\Gamma'(x)}{\Gamma(x)}$, for $n\in\mathbb{N}$ we have $$\Gamma'(n+1) = \Gamma(n+1) \cdot \left(\sum_{k=1}^{n} \frac{1}{k} - \gamma\right) = n! \cdot \left(\sum_{k=1}^{n} \frac{1}{k} - \gamma\right),$$ where $\gamma$ is the Euler–Mascheroni constant. So, $\Gamma'(1) = -\gamma$ and $\Gamma'(2) = 1 - \gamma$.
As $\Gamma'(x)$ is increasing and $\gamma \approx 0.577$, we have $-\gamma \leq \Gamma'(x) \leq 1 - \gamma$ for $x\in[1, 2]$, which leads to $\left|\Gamma'(x)\right| \leq \gamma < 1$ for $x\in [1, 2]$. So, according to the Banach fixed point theorem, $\Gamma(x)$ is a contraction on $[1, 2]$ and has a unique fixed point $x^* \in [1, 2]$ such that $x^* = \Gamma(x^*)$, which can be found by iterating $x_{n+1} = \Gamma(x_n)$ for any $x\in[1, 2]$.
EDIT: just as an interesting fact, it turns out that there are many other regions in the complex plane such that $z_{n+1} = \Gamma(z_n)$ converges to 1: plot for $\mathrm{Re} \; z, \; \mathrm{Im} \; z \in [-5, 5]$ (although it says "absolute value", the limit equals real number 1):
Here is also zoomed out version for $\mathrm{Re} \; z, \mathrm{Im} \; z \in [-20, 20]$, some pattern can be seen for $\mathrm{Re} \; z > 0$.