Proof limit $b^n n^{\alpha} \to 0$, when $n \to \infty$ and $\forall |b| < 1 $. $$\lim_{n \to \infty} b^n n ^{\alpha} = 0, \forall |b| < 1, \alpha > 0$$
I have proved it for cases when $\alpha \le 0$, but I am not sure where to go further. I have tried D'Alembert's test, but I believe that it does not work for $\alpha > 0$. 
Can you give me any direction where to go ? At least, some nominal ideas.

Because the limit of $b^n$ is $0$, since $|b| < 1$ and $n$ approaches infinity
$$\lim_{n \to \infty} (b^n) (n^a) = \lim_{n \to \infty} (b^n) \cdot \lim_{n \to \infty}(n^a)$$
since limit of $b^n$ is $0$, the limit of the previous one is $0$.
