Will this function grow faster than Busy Beavers as $n \to \infty$ Consider the following function: 
$$f(x)=x \uparrow ^{x} x$$
Where the notation $\uparrow$ is Knuth's up-arrow notation and $\uparrow ^{n}$ means $n$ number of up-arrows. For example, $2\uparrow ^{4}3=2\uparrow \uparrow \uparrow \uparrow 3$
Let the Busy Beavers Function be represented by $\Sigma (n)$.
Will there be a threshold value for $n$ such that $f(n) > \Sigma(n)$?
Bonus Question: Assuming that $f(n)$ does surpass $\Sigma(n)$ at some value of $n$, will it be possible to determine if this value would occur before or after $\Sigma(n)$ surpasses $\text{TREE}(n)$?
 A: The busy beaver function grows faster than every computable function : Otherwise we could bound the number of steps a turing machine can do before halting and hence solve the halting problem, which is however impossible. Hence the busy beaver function eventually "outgrows" also $TREE(n)$
We can only guess for which $n$ we have $\Sigma(n)>TREE(n)$, but there must be some $n_0$, such that $\Sigma(n)>TREE(n)$ holds for every $n\ge n_0$. We have no upper bound for $\Sigma(n)$ with $n\ge 5$. $n_0=100$ might be an overkill, but we cannot find it out anyway (unless someone finds a machine with $100$ or less states making more than $TREE(100)$ steps which seems unrealistic to me).
Wythagoras proved $$\Sigma(7)>10\uparrow 10\uparrow 10\uparrow 10\uparrow 10\uparrow 7>>10 \uparrow \uparrow 5$$ which is already insanely large , but even the determination of $\Sigma(6)$ seems unrealistic.
The function $f(n)$ has a growth rate comparable to Ackermann-functions (Just $f_{\omega}(n)$ in the fast growing hierarchy). Hence $f(n)$ with a "reasonable" input $n$ cannot compete with $TREE(3)$ . It cannot even compete with the much smaller conway chains.
