Calculation of no. of real solution of $ x^{x^{2015}} = 2015.$ 
Calculation of no. of real solution of $\displaystyle x^{x^{2015}} = 2015.$

My Attempt :: I have calculated for $x>0$, 
$\bullet \;$ If $0<x\leq 1$, then $\bf{L.H.S}<1$ and $\bf{R.H.S>1}$. So no solution.
$\bullet \;$ If $x>1\;,$ then let $\displaystyle y=x^{x^{2015}}$. Then $\displaystyle \ln(y) = x^{2015}\cdot \ln(x)$
So $\displaystyle \frac{dy}{dx} = x^{x^{2015}}\cdot x^{2014}\cdot \left(1+2015\cdot \ln x\right)\;,$ Now for Maximum and Minimum $\displaystyle \frac{dy}{dx} = 0$
So we get $\displaystyle \ln x=-\frac{1}{2015}\Rightarrow x = e^{-\frac{1}{2015}}.$
Now, how can I proceed further and calculate the solutions for $x<0$?
 A: For solving for the root, it would be more practical to solve for $0$$$f(x)=2015\log(x)+\log\big(\log(x)\big)-\log\big(\log(2015)\big)$$ $$f'(x)=\frac{2015}{x}+\frac{1}{x \log (x)}$$ and use Newton starting at $x_0=1+\epsilon$. Using $\epsilon=10^{-6}$, the following iterates are $1.00002$, $1.00023$, $1.00178$, $1.00364$, $1.00378$ which is the solution for six significant figures.
A: Analytical solving :
$$x^{x^a}=b$$
$$x^a \ln(x)=\ln(b)$$
$$x^a a \ln(x)=a \ln(b)$$
$$x^a\ln(x^a)=a\ln(b)$$
$$\ln(x^a)=\frac{a\ln(b)}{x^a}$$
$$x^a=\exp\left(\frac{a\ln(b)}{x^a}\right)$$
$$a\ln(b)x^a=a\ln(b)\exp\left(\frac{a\ln(b)}{x^a}\right)$$
$$\left(\frac{a\ln(b)}{x^a}\right)\exp\left(\frac{a\ln(b)}{x^a}\right)=a\ln(b)$$
with $X=\frac{a\ln(b)}{x^a}$
$$X e^X=a\ln(b)$$
$$X=W\left(a\ln(b)\right)$$
$W$ is the LambertW function
$$x^a=\frac{a\ln(b)}{W\left(a\ln(b)\right)}$$
$$x=\left(\frac{a\ln(b)}{W\left(a\ln(b)\right)}\right)^{1/a}$$
Numerical computation in case of $a=b=2015$ :
$$x=\left(\frac{2015\ln(2015)}{W\left(2015\ln(2015)\right)}\right)^{1/2015}\simeq 1.0037830058$$
Of course, the approximate can be computed more directly thanks to recursive methods of numerical solving.
A: since the function $x^{x^{2015}}$ is increasing and goes to infinity, since $f(1)=1$, there is one and only one solution to the equation. since you know that $2^{2^{2015}} > 2015$, you know that the solution is between 1 and 2.
as for the solution, a computer manual calculation shows that 1.0037830057966099 is a very close approximation, as $$1.0037830057966099^{1.0037830057966099^{2015}}=2015.000000025717$$
