An algebra problem by Priyanshu Mishra

We note that $\dfrac d{dx} x^{x^x} = x^{x^x} \left(\dfrac d{dx} x^x \ln x \right)$ $= x^{x^x} \left(x^x (\ln x +1)\ln x + x^{x-1} \right)$ $= x^{x^x + x-1} \left(x\ln^2 x +x + 1 \right) > 0$ for $x > 0$ . This means that $x^{x^x}$ is strictly increasing function $\in (0, \infty)$ for $x \in (0, \infty)$ . Therefore, there is only $\boxed{1}$ solution for $x^{x^x}= 1996$ at $x \approx 2.425$ .