Some tetration you got there

It's trivial that $\displaystyle \lim_{x \to 0} f(x;1) = 0$ , apply L'hopital rule in the following limit

$\displaystyle y = \lim_{x \to 0} f(x;2) \Rightarrow \ln y = \lim_{x \to 0} x \ln x = \lim_{x \to 0} \frac { \ln x}{1/x} = \lim_{x \to 0} \frac{ 1/x}{-1/x^2} = \lim_{x \to 0} -x = 0 \Rightarrow y = e^0 = 1$ , it's obvious that the left limit and right limit are equal so the limit is defined.

I claim that $\displaystyle \lim_{x \to 0} f(x;n)$ yields $0$ and $1$ for odd and even positive integers respectively.

For odd numbers, prove by Mathematical Induction, we have the basic step which is mentioned above.

Inductive step: let $k$ be a positive odd number such that

$\displaystyle \lim_{x \to 0} f(x;k) = 0$ is true

Then $k+2$ is the smallest positive odd number bigger than $k$ , denote $M$ as such

$\LARGE \displaystyle M = \lim_{x \to 0} f(x;k+2) = \lim_{x \to 0} x^{x^{f(x;k)} } = \lim_{x \to 0} x^{x^0} = 0$ which is also true. Thus for all positive odd numbers $M$ , $\displaystyle \lim_{x \to 0} f(x;M) = 0$

By using an analogous argument, we have for all positive even numbers $N$ , $\displaystyle \lim_{x \to 0} f(x;N) = 1$

Because there's the same number of odd numbers and even numbers in the range of $1$ to $10000$ inclusive. $p = \frac {1}{2} \Rightarrow \lfloor 1000p \rfloor = \boxed{500}$

It is not hard to see that the given limit is $1$ for even $n$ and $0$ for odd $n$ . So $p$ is simply the probability of selecting an even number from $1,2,..., 10000$ , which is $0.5$ . So the answer to the question is $500$ .