Too many ceilings

The best way to solve this problem is to break $x$ into its components .

Let $x=\color{#20A900}I+\color{#624F41}f$ where $\color{#20A900}I$ is the Integral part of $x$ and $\color{#624F41}f$ is the fractional part of $x$ such That $0<\color{#624F41}f<1$ . Now ,

$\lceil x \rceil = \lceil \color{#20A900}I+\color{#624F41}f \rceil = \color{#20A900}I+1 \\ \{ x \} = \{ \color{#20A900}I+\color{#624F41}f \} = \color{#624F41}f \\ \lceil \{ x \} \rceil = \lceil \color{#624F41}f \rceil = 1 \\ \{ \lceil x \rceil \}= \{ \color{#20A900}I+1 \} = 0 \\ \text{Putting the Values in the expression , we get } \\ \implies ( \lceil ( \lceil(0+1+\color{#624F41}f) \rceil +\color{#624F41}f) \rceil)! \\ \\ \implies ( \lceil ( \lceil 1+\color{#624F41}f \rceil +\color{#624F41}f) \rceil)! \\ \implies ( \lceil (2 +\color{#624F41}f) \rceil)! \implies (3)! = \color{#3D99F6}{\boxed{6}}$

$\left \lceil x \right \rceil$ is always an integer and fracrional function of any integer is always 0. So $\{ \left \lceil x \right \rceil \}$ = 0.
Now, $\{x\}$ always lies between 0 and 1. It can be 0 when it is an integer but x being irrational can never give its fractional function as 0. So for any values lying between 0 and 1, the floor function is always 1. So, $\left \lceil \{x\} \right \rceil$ = 1.

Now, $\{x\}$ is always less than 1 but greater than 0 for any irrational number as explained above. So, its value is 0.something.
So, evaluating the inner bracket, $\{ \left \lceil x \right \rceil \} + \left \lceil \{x\} \right \rceil + \{x\}$ = 1.something. Now, the floor of this value is 2. So, $\left \lceil \left( \{ \left \lceil x \right \rceil \} + \left \lceil \{x\} \right \rceil + \{x\} \right) \right \rceil = 2$ .

Again, $\{x\}$ is always less than 1 but greater than 0 for any irrational number as explained above. So, its value is 0.something.
So, $\left \lceil \left( \{ \left \lceil x \right \rceil \} + \left \lceil \{x\} \right \rceil + \{x\} \right) \right \rceil + \{x\}$ = 2.something.

So, $\left \lceil \left( \left \lceil \left( \{ \left \lceil x \right \rceil \} + \left \lceil \{x\} \right \rceil + \{x\} \right) \right \rceil + \{x\} \right) \right \rceil = 3$ .

So, the answer which we want is the factorial of the above expression which is 3! = $\boxed{6}$ .

What I did was plug in pi and derive the answer from there. But there are fancy explanations that tells you why it works above and/or below me!