In trying to give the OP an elementary answer to this question, I made some rather stupid mistakes. I feel terrible about giving a wrong answer (in lieu of a complicated but correct one).
I devised a new proof, and wanted to check it before editing my answer. Does everyone like the following (well enough)?
Assertion: $\lim_{x \to 0^+} \frac{x^{x^x}}{x} = 1$
Proof: We pass to the log of the limit.
$\log\left(\lim_{x \to 0^+} \frac{x^{x^x}}{x}\right) = \lim_{x \to 0^+} \log\left(\frac{x^{x^x}}{x}\right) = \lim_{x \to 0^+} \frac{\log(x)}{\frac{1}{x^x - 1}}$
We use L'Hospital's rule, and rearrange:
$\lim_{x \to 0^+} \frac{\log(x)}{\frac{1}{x^x - 1}} = \lim_{x \to 0^+} \frac{\frac{1}{x}}{\frac{-x^x(\log(x) + 1)}{(x^x - 1)^2}} = \lim_{x \to 0^+} \frac{- (x^x - 1)^2}{x^{x}(x\log(x) + x)} = \left( \lim_{x \to 0^+} \frac{-(x^x - 1)}{x^x} \right) \left( \lim_{x \to 0^+} \frac{(x^x - 1)}{x\log(x) + x} \right) $
provided that both of these last limits exist; but (again using L'Hospital in the 2nd limit) we see that
$\lim_{x \to 0^+} \frac{-(x^x - 1)}{x^x} = \frac{0}{1} = 0,$
$\lim_{x \to 0^+} \frac{(x^x - 1)}{x\log(x) + x} = \lim_{x \to 0^+} \frac{x^x(\log(x)+1)}{(1+ \log(x)) + (1)} = \left( \lim_{x \to 0^+} \frac{x^x(\log(x)+2)}{( \log(x)) + 2)} - \lim_{x \to 0^+} \frac{x^x}{(\log(x)) + 2)} \right) = 1,$
and therefore $\displaystyle\log\left(\lim_{x \to 0^+} \frac{x^{x^x}}{x}\right) = 0$.
Evaluating both sides by $\exp(x)$ therefore shows that $\displaystyle\lim_{x \to 0^+} \frac{x^{x^x}}{x} = 1$.