prove $\lim_ {n \rightarrow \infty} \frac{b^n}{n^k}=\infty$ 
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How to prove that exponential grows faster than polynomial? 

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.
 A: Note that $$b^n = \exp(n \log b) = \displaystyle \sum_{j=0}^{\infty} \dfrac{(n \log b)^j}{j!} \geq \dfrac{(n \log b)^{k+1}}{(k+1)!}$$
Hence, we have
$$\dfrac{b^n}{n^k} \geq \dfrac{(n \log b)^{k+1}}{(k+1)! n^k} = \dfrac{(\log b)^{k+1}}{(k+1)!} n$$
Hence, $$\lim_{n \to \infty} \dfrac{b^n}{n^k} \geq \lim_{n \to \infty} \dfrac{(\log b)^{k+1}}{(k+1)!} n = \infty$$
A: $$\lim_{n \rightarrow \infty} \frac{b^n}{n^k}=\infty$$
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.

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.
A: We have $b^n/n^k=\exp(n\ln b-k\ln n)=\exp(n\ln b(1-\frac{k}{\ln b}\frac{\ln n}{n}))$.
Now $\lim_{n\rightarrow +\infty} n\ln b=+\infty$ and $\lim_{n\rightarrow +\infty} 1-\frac{k}{\ln b}\frac{\ln n}{n}=1$.
So $\lim_{n\rightarrow +\infty}n\ln b(1-\frac{k}{\ln b}\frac{\ln n}{n})=+\infty$.
It follows that $\lim_{n\rightarrow +\infty} b^n/n^k=+\infty$.
A: This follows immediately from Bernoulli:
$$\frac{b^n}{n^k}= (\frac{b^{\frac{n}{2k}}}{\sqrt{n}})^{2k}$$
Now, by Bernoully
$$b^{\frac{n}{2k}} \geq 1+\frac{n}{2k}(b-1)$$
Thus
$$\frac{b^n}{n^k} \geq  (\frac{1}{\sqrt{n}}+\frac{\sqrt{n}}{2k}(b-1))^{2k} \geq (\frac{\sqrt{n}}{2k}(b-1))^{2k}$$
Since th RHS goes to infinity, you are done.
A: Apply L'Hospital's Rule $k$ times to get $$\frac{(\log b)^k b^n}{k!}$$ if $k>0$
