An iteration formula I found (please don't jump at me if it's already been discovered) $\sin^2 \theta$">
Where the modulus-like symbol actually denotes iteration of a radical function. Sorry for the messy work everyone- I am new to this stuff, and I literally just found this iteration formula like 15 minutes ago. $X[i]$ starts as being an arbitrary number greater than $n!/(n-2)!$ which I should have mentioned earlier but didn't have space to (apologies).
So the business is: you start with a number e.g. 1, then if you want to find say $50$, you compute $sqrt(1 + 50!/48!)$ then replace $1$ on your calculator with the result you just got a.k.a. $Ans$ etc. and keep iterating till you get (at a surprisingly fast convergence rate!) the very number you started with, n. 
I appreciate this might not be the most interesting insight or $n!/(n-2)!$ is probably just me not realising that it could also be presented as $n(n-1) = n^2-n$. But still, please don't mass-downvote it all, remember it's a 15-year old person who figured this out...
 A: You can do this formally by just looking at
$$\sqrt{n!/(n-2)!+\sqrt{n!/(n-2)! + \dots}}$$
and so on to infinity. Suppose this has a limit. Then we'd expect that
$$L=\sqrt{n!/(n-2)! + L}$$
This gives
$$L^2-L=n!/(n-2)!=n^2-n$$
which has the obvious solution of $L=n$.
Analytically, this could be done with some sort of fixpoint theorem (such as the Banach fixed-point theorem), setting
$$f(x)=\sqrt{n!/(n-2)! + x}$$
and using it to show that $\lim_{m\to \infty}\underbrace{f(\cdots f(x)\cdots)}_{m\text{ times}}$ is the solution to the fixpoint equation
$$f(L)=L.$$
A: So you're saying that the sequence
$$X_0=1$$
$$X_{k+1}=\sqrt{X_k+\frac{n!}{(n-2)!}}$$
tends to towards $n$. Firstly notice that
$$\frac{n!}{(n-2)!}=n(n-1)=n^2-n$$
Hence your sequence is equivalent to
$$X_{k+1}=\sqrt{X_k+n^2-n}$$
Then note that for $0\le X_k\lt n$ we have
$$X_k\lt\sqrt{X_k+n^2-n}\lt n$$
So your sequence is strictly increasing and bounded above. Also note that
$$\sqrt{n+n^2-n}=\sqrt{n^2}=n$$
Hence the sequence has a stationary point at $X_k=n$. This is sufficient to prove that for a given initial value $0\le X_0\le n$ the sequence tends to $n$. In other words
$$\lim_{k\to\infty}X_k=n$$
In fact one can prove that this sequence tends to $n$ for any initial value by also using the fact that $\sqrt{X_k+n^2-n}\lt X_k$ when $X_k\gt n$.
