A number to its own power

Minimise the logarithm of $f$ instead; it has the same minimiser.

1. Write $\ln f(x)=x\ln x$, which has derivative $\ln x+1$.
2. The derivative is zero at $x=\frac{1}{e}$, negative before it and positive after it, so this is the minimum.
3. There $\ln f=-\frac{1}{e}$, so the minimum value is $e^{-\frac{1}{e}}$.
4. Since $\frac{1}{e}\approx 0.36788$, the minimum is $e^{-0.36788}\approx 0.692$.

Answer: $\approx 0.692$