define $ a_{n}=\frac{(1-\alpha^{k})^{n}}{(1-\alpha^{n})^{k}}$ $ 0 <\alpha < 1$ $ n,k$ being positive integers. clearly $ a_{n}$ is decreasing in $ n$ . observe $ a_{k}= 1$ so for $ n > k$ $ a_{n}< 1$ putting $ \alpha =\frac{y}{x}$ $ y < x$ we get $ (x^{k}-y^{k})^{n}< (x^{n}-y^{n})^{k}$

Put $ a=\frac{x}{y}>1$ , we need prove $ (a^{k}-1)^{n}<(a^{n}-1)^{k}$ or $ \frac{\ln (a^{k}-1)}{k}<\frac{\ln (a^{n}-1)}{n}$. This is true because $ f(t)=\frac{\ln (a^{t}-1)}{t}$ is increasing on $ (0,\infty)(f'(t)>0\forall t>0)$.