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> Calculation of no. of real solution of $\displaystyle x^{x^{2015}} = 2015.$
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**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 $x<0$?