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.
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 $$e^{f(n)}\not\sim e^{g(n)}.$$