Can we provide a good estimation for $(n!)!$? I was thinking about this

$$(n!)!$$

for $n\in\mathbb{N}$.
I wanted to find a suitable approximation, or in any case a very good estimation for this. My first idea was to use Stirling approximation for $(n!)!$ but it leads me to a very confusing algebra. Also I tried to make some approximations but I came up with a negative form.
Is there something we can do to obtain a quite good trend for this? 
If you need details of what I wrote, or about what I wrote down on the paper, tell me. It's just a pretty chaotic mess so I avoided to write it down here for the sake of simplicity.
 A: The bad news is that that nasty algebra does not necessarily give a valid approximation, at least not an approximation of the sort denoted by $\sim$.
Let's say $$S(n)=\sqrt{2\pi n}\left(\frac ne\right)^n,$$so Stirling's formula says that $$n!\sim S(n)\quad(n\to\infty).$$It follows that $$(n!)!\sim S(n!),$$but it does not follow, at least not immediately, that $$(n!)!\sim S(S(n)).$$
Why not? Saying $n!\sim S(n)$ means that $$\frac{n!}{S(n)}\to1.$$That certainly implies $$\frac{(n!)!}{S(n!)}\to1,$$but there's no reason to think that it implies that $$\frac{(n!)!}{S(S(n))}\to1.$$
For example, say $f(n)=n^2$ and $g(n)=n^2+n$. Then it's clear that $$f(n)\sim g(n)\quad(n\to\infty),$$but $$\frac{e^{f(n)}}{e^{g(n)}}=e^{-n}\not\to1,$$so $$e^{f(n)}\not\sim e^{g(n)}.$$
A: An easy but bad approximation
for $n!$
is
$(n/e)^n$.
Plugging this into Stirling
in two ways,
$(n!)!
\approx \sqrt{2\pi  (n/e)^n}((n/e)^n/e)^{(n/e)^n}
$
or
$(n!)!
\approx \left(\dfrac{\sqrt{2\pi n}(n/e)^n}{e}\right)
^{\sqrt{2\pi n}(n/e)^n}
$
where
"$\approx$" means
"somewhere in the neighborhood
but nobody really cares
since the numbers
are so damn large
for large $n$
that we really ought to take the log
anyway."
