Is there official notation to represent "perform an operation n times"? I would like to know if you can represent the idea of say,
$$n^n$$
n amount of times without defining a function.  For example:
$$1$$
$$2^2$$
$$3^{3^3}$$
$$4^{4^{4^4}}$$
$$5^{5^{5^{5^5}}}$$
and so on, n amount of times.  Is there a way to represent this mathematically without defining a function?
EDIT: I understand that in this particular example the answer is tetration.  However, I am searching for more general notation that could also work with the following:
$$\ln(n)$$
$$n!$$
$${^n}n$$
all n amount of times.  Another example with factorial:
$$1!$$
$$(2!)!$$
$$((3!)!)!$$
$$(((4!)!)!)!$$
$$((((5!)!)!)!)!$$
 A: It sounds like you are looking for the notation for an iterated function.  If $f$ is the function you want to iterate, you generally write
$$f^n$$
for the $n^\text{th}$ iterate of $f$.  (Notice that this is different from the notation for the $n^\text{th}$ derivative of $f$, which is given by $f^{(n)}$).  So, for example, you  could write
$$f(n) = n!$$
and then
$$f^1(n) = n!$$
$$f^2(n) = (n!)!$$
and so on.
A: To add to Strants' answer, which indeed gives the most common way of writing "do $f$ to an argument $n$ times", I'll give the "expanded form" of this which is $$f^n(x) \equiv f(f(\stackrel{(n)}{\cdots}f(x))$$
This is cumbersome and I strongly recommend using the compact notation, even if it means defining a new function. However it might suit your purposes for cases where the function has a reserved definition for the superscript (e.g. $\sin^2$), or you don't want to introduce a new function name. So for example, $$\sin(\sin(\stackrel{(n)}{\cdots}\sin(x)))$$
would apply the sine function $n$ times to $x$. However, take for example $$f(x) = \frac{2x+1}{x^2+e^x}$$
The $n$-iterated expression for it is (for me) unimaginable, and even if presented with it I would most likely not know at all what it was meant to designate. Of course, I could always write $f^n(x)$ and not have a doubt in the world :)
A: You can use Knuth's up-arrow notation:
$a\uparrow n=a^n$
$a\uparrow \uparrow n=a^{a^{a^{\dots}}} $ n times
$a\uparrow \uparrow \uparrow n= a\uparrow \uparrow a\uparrow \uparrow a\uparrow \uparrow \dots$ n times
and in general
$a\uparrow ^b n=a\uparrow ^{b-1} a$ iterated n times
