They look similar...(4)

$\begin{aligned} L & = \lim_{x \to 0^+} \left(x^{x^x} - (x^x)^x\right) \\ & = \lim_{x \to 0^+} x^{x^x} - \lim_{x \to 0^+} (x^x)^x \\ & = \lim_{x \to 0^+} \exp (x^x \ln x) - \lim_{x \to 0^+} \exp (x \ln (x^x)) & \small \color{#3D99F6} \text{where }\exp (x) = e^x \\ & = \exp \left({\color{#3D99F6}\lim_{x \to 0^+} x^x} \lim_{x \to 0^+} \ln x \right) - \exp \left( \lim_{x \to 0^+} x \ln \left( \color{#3D99F6} \lim_{x \to 0^+} x^x\right)\right) & \small \color{#3D99F6} \text{See note } \lim_{x \to 0^+} x^x = 1 \\ & = e^{\color{#3D99F6}1} e^{-\infty} - e^0 e^{\ln \color{#3D99F6}1} = 0 - 1 = \boxed{-1} \end{aligned}$

---

Note:

$\begin{aligned} \lim_{x \to 0^+} x^x & = \lim_{x \to 0^+} \exp (x \ln x) \\ & = \lim_{x \to 0^+} \exp \left(\frac {\ln x}{\frac 1x} \right) & \small \color{#3D99F6} \text{A }\infty/\infty \text{ case, L'Hôpital's rule applies.} \\ & = \lim_{x \to 0^+} \exp \left(\frac {\frac 1x}{-\frac 1{x^2}} \right) & \small \color{#3D99F6} \text{Differentiate up and down w.r.t. }x \\ & = \lim_{x \to 0^+} \exp \left(-x\right) \\ & = e^0 = 1 \end{aligned}$