Let $$ f(x)=x^{2025}-x^{2000}, x \text{ }\epsilon \text{ }[0,1] $$ and the minimmu value of the function $$ f(x)$$ in the interval [0, 1] be $$(80)^{80}(n)^{-81}$$. Then n is equal to

$$f'(x) = 2025x^{2024} - 2000x^{1999} = x^{1999}(2025x^{25} - 2000)$$

critical point: Set $$f'(x) = 0 \implies x^{25} = \frac{2000}{2025} = \frac{80}{81} \implies x = \left(\frac{80}{81}\right)^{1/25}$$

$$f(x_{\min})$$:

$$f(x) = x^{2000}(x^{25} - 1)$$

$$f(x_{\min}) = \left(\frac{80}{81}\right)^{\frac{2000}{25}} \left(\frac{80}{81} - 1\right) = \left(\frac{80}{81}\right)^{80} \left(-\frac{1}{81}\right) = -(80)^{80} \cdot (81)^{-81}$$

Comparing with $$(80)^{80}(n)^{-81}$$, we get n = -81.