The fraction exceeding its n^th power by the greatest possible number, where n ≥ 2, is

$\left(\frac{1}{n}\right)^{\frac{1}{n-1}}$

Let x be the fraction and let $y=x-x^n$. Then,

$\frac{d y}{d x}=1-n x^{n-1}$ and $\frac{d^2 y}{d x^2}=-n(n-1) x^{n-2}$

Now, $\frac{d y}{d x}=0 \Rightarrow x=\left(\frac{1}{n}\right)^{\frac{1}{n-1}}$

Clearly, $\frac{d^2 y}{d x^2}<0$ for $x=\left(\frac{1}{n}\right)^{\frac{1}{n-1}}$

Hence, y is maximum when $x=\left(\frac{1}{n}\right)^{n-1}$