The interval in which the function $$f(x) = x^x, x > 0$$, is strictly increasing is

Differentiate the function.

To differentiate $$y = x^x$$, use logarithmic differentiation:

$$\ln y = x \ln x$$

$$\frac{1}{y} \frac{dy}{dx} = \ln x + x(\frac{1}{x}) = \ln x + 1$$

$$f'(x) = x^x (1 + \ln x)$$

Step 2: Determine where $$f'(x) > 0$$.

Since $$x^x$$ is always positive for $$x > 0$$, the sign depends on $$(1 + \ln x)$$.

$$1 + \ln x > 0$$

$$\ln x > -1$$

$$x > e^{-1} \implies x > \frac{1}{e}$$

Thus, the function is strictly increasing on the interval $$[1/e, \infty)$$.

Correct Option: C ($$[1/e, \infty)$$)