How to solve for $y$ the equation $x= y^y$? I need an equation where I receive a number that when raised to itself equals the input.
Formally: in $x=y^y$ solve for $y$.
Intro to Calculus level knowledge. If the Lambert function is necessary, please explain it to me. 
 A: For sure, the most elegant way corresponds to the use of Lambert function and get $$y=e^{W(\log x)}$$ But, let us suppose that you cannot access this function. The Wikipedia page gives approximation $$W(a)\approx L_1-L_2+\frac{L_2}{L_1}+\frac{L_2(L_2-2)}{2L_1^2}+\frac{L_2(2L_2^2-9L_2+6)}{6L_1^3}+\cdots$$ where $L_1=\log(a)$ and $L_2=\log(L_1)$. This would provide a very good estimate $y_0$. Let us try with $x=10^6$; the above formula would give $W\big(\log(10^6)\big)=1.9483$ and then $y=7.01678$ while the exact value is $y=7.0658$.
Having such an estimate, Newton method could be used. For better conditionning, I suggest to write the equation as $$f(y)=y \log(y)-\log(x)=0$$ $$f'(y)=1+\log(y)$$ and the iterative scheme write $$y_{n+1}=\frac{\log (x)+y_n}{\log (y_n)+1}$$ Starting from $y_0=7.0658$, the successive iterates would be $7.0657967282998769173$, $7.0657967282996206111$,  which is the solution for $20$ significant digits
A: Yes, the Lambert W function
will do what you want.
Its defining relation is
$z = W(z)e^{W(z)}$.
Here is one of many places
it is discussed:
http://en.wikipedia.org/wiki/Lambert_W_function
In particular,
to solve
$x^x = z$,
set
$x = e^{W(\ln z)}$
or equivalently,
$x
=\dfrac{\ln z}{W(\ln z)}
$.
This is example 2 in
the previous link.
A: $W(x)$ is defined as the inverse of $xe^x$. Therefore, we have to do some manupulations in order to reach the form to apply theW function.
$$x=y^y$$
$$\log x=y\log y$$
$$\log x=\log y\cdot e^{\log y}$$ 
$$W(\log x)=W(\log y \cdot e^{\log y})=\log y$$
$$\log y=W(\log x)$$
$$y=e^{W(\log x)}$$
