Math Help - Convergence of a series

1. Convergence of a series

Hi everyone.
I need to prove that if 0 < a < 1 , then the series $\sum_{n=1}^{\infty} (1-\frac{1}{n^a})^n$ converges.
I tried to use the limit test and the comparison test but I couldn't manage to.
Can someone give me a hint how to prove that?

I need to prove that if 0 < a < 1 , then the series $\sum_{n=1}^{\infty} (1-\frac{1}{n^a})^n$ converges.
Notice that $\left(1-\frac{1}{n^\alpha}\right)^n=e^{n\ln\left(1-\frac{1}{n^\alpha}\right)}$. But, notice that since $0<\frac{1}{n}\leqslant 1$ we may apply the series to the natural log to obtain that $e^{n\ln\left(1-\frac{1}{n^\alpha}\right)}=e^{-n\left(\frac{1}{n^\alpha}+\frac{1}{2n^{2\alpha}}+\ cdots\right)}\approx e^{-n^{1-\alpha}}$ so what happens if $0<\alpha<1$?