First of all, we need to see how to define this function and if it exists.
The truth is that there are two functions behind it, not one.

Define the family of functions:

$$f_n(x)=x^{(n+1)f_{n+1}(x)}$$

But, there is an instant trouble. First of all we have

$$f_1(1)=1$$ and
$$f_1(1^{+})=\infty$$

The trouble is how to define $f(0)$. Let us define a partial function

$$
f_{m,n}(x)=\begin{cases}
& x^{f_{m,n+1}(x)},\text{ if } n \leq m \\
& 1,\text{ if } n>m
\end{cases}
$$

and then try to find the limiting function. And here is the problem.

If $m$ is even then the function starts at $1$ and has a jump at $\frac{1}{m}$ towards $0$.
If $m$ is odd then the the function starts at $0$ and has a jump at $\frac{1}{m}$ towards $1$.

Going towards infinity, we will obtain two versions of the functions: *even* that starts at $0$ and is monotonically increasing and *odd* that has a minimum between $0$ and $1$ and starts at $1$.

(We can agree that neither of two is continuous at $0$ or we need to extend the function at $0$ ourselves, because the limiting process is never going to reach $0$ or $1$ value at $f_1(0)$ by itself.)

odd and even case

odd and even case

Now regarding first derivative, there is no closed form of it. We can simply state it the same way the function is defined and then use an infinite chain to define it in full.

$$f_n'(x)=(n+1)x^{(n+1)f_{n+1}(x)-1} \left (x\ln(x)f'_{n+1}(x)+f_{n+1}(x) \right )$$

This makes it obvious that $f_1'(1)=2 \cdot 3 \cdot 4 \cdot 5 ... = \infty$ while $f_1'(0) = 0$

Now one thing is certain and that is that the difference between $f_{2n}$ and $f_{2n+2}$ is all smaller and smaller as $n$ tends infinity. The same for $f_{2n+1}$ and $f_{2n+3}$.

so we can try to use that and define that at $n \gg 1$

$$f_n(x)=y \sim x^{(n+1)x^{(n+2)y}}$$

Now, here comes the story. First, since $x$ is between $0$ and $1$ $x^n$ for very large $n$ tends to $0$. So we arrive at the expression

$$y = 0^{0^y}$$

which can lead only to $y=0$ or $y=1$

So at infinity we have $f_n(x)$ reaching flat $1$. (We could take $0$ as well but that is the same as saying that $f_{n-1}(x)$ reaches $1$. So $1$ is more universal.)

For this reason, our analysis of starting with the above definition for $f_{m,n}(x)$ is just right.

Other than this it is not likely that it is possible to write this function in any known and simple closed form even with all other special functions available. It is simply a beast on its own starting from its dual nature.

Another curious question is how to represent these exponential functions. It is difficult to find a fine cut and the difference between

$$x^{2^{x^{3}}}$$

and

$$x^{2x^{3}}$$

and

$$x^{(2x)^{3}}$$

This notation should reduce ambiguity and do notice that extended exponential is no longer commutative.

$$x^{2 \cdot x^{3 \cdot x^{4 \cdot x^{5 \cdot x...}}}}=\operatorname*{E}\limits_{n=1}^{\infty}x^{n+1}$$

$$x^{(2x)^{(3x)^{(4x)^{(5x)...}}}}=\operatorname*{E}\limits_{n=1}^{\infty}nx$$

$x^{2^{x^{3}}}$ is obviously a special case with alteration
$$
(x,n) = \begin{cases}
& x, n \mod 2=1\\
& \frac{n}{2}+1, n \mod 2=0
\end{cases}
$$

$$\operatorname*{E}\limits_{n=1}^{\infty}(x,n)$$

A general problem with this extended exponentiation is that you technically need to read it right to left in order to clear the confusion.