What is a better way to prove the identity: $\frac{1}{x} = \log_{x^x}(x)$ where $x>0$

Question

This is my attempt at an algebraic proof of the identity $\frac{1}{x} = \log_{x^x}(x)$ where $x>0$: $$ \text{Let} \space n = \log_{x^x}(x)\space \text{where} \space x>0 $$ $$\log_{x^x}(x)=n$$ $$(x^x)^n=x$$ $$\ln((x^x)^n)=\ln(x)$$ $$xn\cdot \ln(x)=\ln(x)$$ $$xn=1$$ $$n=\frac{1}{x}$$