Growth of $n!!\dots !$ The asymptotic growth of the factorial function $n!$ is famously given by Stirling's formula as
$$n! \sim \sqrt{2 \pi n}\left(\frac{n}{e}\right)^n$$
Is there a similar formula for the iterated factorial
$$n\underbrace{!!\dots !}_n$$
perhaps in terms of tetration ${^{n}n}$ ?
The function grows fast to say the least, with the first terms being
$$
\begin{align}
1! &= 1 \\
(2!)! &= 2 \\
((3!)!)! &\approx 2.6\times 10^{1746}
\end{align}
$$
and all following terms too large even for scientific notation.
 A: Let $_k n$ denote $n!!\cdots !$ with $k$ factorials.
We will show that, given $n \ge 3$ and $k \ge 1$, $_k n$ lies between $^kn$ and $^{k+1} n$. 
The key inequalities that we will make use of are
$(n/e)^n < n! < n^{n-1}$  for $n \ge 3$.
The left inequality follows from a Riemann Sum argument.  The right inequality is obvious from $n * (n-1) \cdots * 2 < n * n \cdots n$.
Part I: Given $n \ge 3$, $_k n > \ ^kn$ 
Proof by induction:
Clearly true for $k=1$.
For $k=2$, $_2 n = n!! \ge (2n)! > n^n = \ ^2n$.
Now, assume $_k n > \ ^kn$ for some $k \ge 2$. Then
$_{k+1} n = (_k n)! > (^k n)! > (\frac{^k n}{e})^{^k n} > n^{^k n} = \ ^{k+1}n$.
Part II: Given $n \ge 3$, $_k n < {^{k+1} n}$.
We will prove the following lemma:
Lemma: Given $n \ge 3$, $_k n < \frac{^{k+1} n}{^k n}$
Proof by induction:
For $k=1$, we have $_1 n = n! < n^{n-1} = \frac{^2 n}{^1 n}$.
Given $_k n < \frac{^{k+1} n}{^kn}$, we have
$_{k+1} n = (_k n)! < (\frac{^{k+1} n}{^kn})! < (\frac{^{k+1} n}{^kn})^{\frac{^{k+1} n}{^kn}} = \frac{(^{k+1} n)^{\frac{^{k+1} n}{^k n}}}{(^{k} n)^{\frac{^{k+1} n}{^k n}}} < \frac{(n^{^k n})^{\frac{^{k+1} n}{^k n}}}{n^{\frac{^{k+1} n}{^k n}}} < \frac{n ^{^{k+1} n}}{n^{^k n}} = \ \frac{^{k+2} n}{^{k+1} n}$.
Therefore:
For $n \ge 3$ and $k \ge 1$, $^k n < _k n < {^{k+1} n}$.
