sequences and series - prove $lim_ {n rightarrow infty} frac{b^n}{n^k}=infty$

I have this sequence with $b>1$ and $k$ a natural, which diverges:
$$\lim_{n \rightarrow \infty} \frac{b^n}{n^k}=\infty$$
I need to prove this, with what i have learnt till now from my textbook, my simple step is this:

Since $n^2\leq2^n$ for $n>3$, i said $b^n\geq n^k$, so it diverges. Is it right?

I am asking here not just to get the right answer, but to learn more wonderful steps and properties.

Answer

$$\lim_{n \rightarrow \infty} \frac{b^n}{n^k}=\infty$$

You can use the root test, too: $$\lim_{ n\to \infty}\sqrt[\large n]{\frac{b^n}{n^k}} = b>1$$

Therefore, the limit diverges.

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The root test takes the $\lim$ of the $n$-th root of the term: $$\lim_{n \to \infty} \sqrt[\large n]{|a_n|} = \alpha.$$

If $\alpha < 1$ the sum/limit converges.

If $\alpha > 1$ the sum/limit diverges.

If $\alpha = 1$, the root test is inconclusive.