Dividing both the denominator and the numerator by $n^{\beta}$ and then expanding the denominator of the expression gives

$$\frac{n^{\alpha}}{n^{\beta}-(n-1)^{\beta}}=\frac{n^{\alpha-\beta}}{1-\left(1-\frac{1}{n}\right)^{\beta}}=\frac{n^{\alpha-\beta}}{1-\left(1-\frac{\beta}{n}+O(\frac{1}{n})\right)}=\frac{n^{\alpha-\beta+1}}{\beta-n\cdot O(\frac{1}{n})}$$

Because

$$\lim_{n\to\infty}\left(n\cdot O\left(\frac{1}{n}\right)\right)=0$$

therefore,

$$\lim_{n\to\infty}\frac{n^{\alpha}}{\beta-(n-1)^{\beta}} = \left\{\begin{array}{ll} 0 &, \alpha-\beta + 1 < 0\\ \frac{1}{\beta}&, \alpha - \beta + 1 = 0 \\ \infty&, \alpha-\beta +1 > 0 \end{array}\right.$$

Comparing this result with the given conditions leads to the conclusion that

$$\beta = \boxed{\frac{1}{2020}}\implies \alpha=\boxed{-\frac{2019}{2020}}$$