How can I prove that $\log^k(n) = O(n^\epsilon)$? How can I prove that $\log^k(n) = O(n^\epsilon)$?
(for any $\epsilon > 0 $ and for any integer and positive $k$).
I tried to do it by - definition , and to prove that exists $c,n_0 > 0$ so that for any $n > n_0 $ exists: $c \cdot n^\epsilon > \log^k(n) $, but I don't have idea how to continue for this point.  
I will be happy to listen ideas :).
 A: By definition we have
$$\log^k(n) = n^\epsilon\cdot \frac{\log^k(n)}{n^\epsilon}$$
with
$$\frac{\log^k(n)}{n^\epsilon}\to 0$$
(refer to Evaluation $\lim_{n\to \infty}\frac{{\log^k n}}{n^{\epsilon}}$)
therefore
$$\log^k(n) = o(n^\epsilon)$$
A: Consider $\;\log\biggl(\dfrac{\log^kn}{n^ε}\biggr)$:
$$\log\biggl(\dfrac{\log^kn}{n^ε}\biggr)=k\log(\log n)-ε\log n=-ε\log n\biggl(1-\frac kε\underbrace{\frac{\log(\log n)}{\log n}}_{\stackrel{\downarrow}{0}}\biggr)$$
so the log tends to $-\infty$ when $n$ tends to $\infty$, i.e.
$$\lim_{n\to\infty}\dfrac{\log^kn}{n^ε}=0\iff\log^kn=o(n^ε),$$
and a fortiori, it is $\:O(n^ε)$.
A: It is well known that $\log (m)<m$ for all $m>0$. Taking $m=n^{\epsilon/k}$ gives
$$\dfrac{\epsilon}{k} \log(n)<n^{\dfrac{\epsilon}{k}}$$
and manipulating this equation gives
$$\log^k (n)<\left(\frac{k}{\epsilon} \right)^kn^\epsilon. $$
A: Here we'll try to come up with an explicit lower bound for $n$ instead of using limits. We need to solve $\log^k(n) \leq n^\epsilon$ (Here I assume the big-Oh constant $c=1$). Assuming $n$ large, in particular $n > 1$ we can rewrite $n = e^m$. for some $m > 0$. This gives us to solve:
$$
\log^k(n) \leq n^\epsilon \Leftrightarrow m^k \leq e^{\epsilon m}
$$ 
Now for $\epsilon, m > 0$, we have
$$
e^{\epsilon m} = \sum_{i=0}^\infty \frac{(\epsilon m)^i}{i!} > \frac{(\epsilon m)^{k+1}}{(k+1)!}
$$
This gives us the implication:
$$
m^k \leq \frac{(\epsilon m)^{k+1}}{(k+1)!} \implies  m^k \leq e^{\epsilon m}
$$
Hence if we manage to solve for $m$ the LHS of the implication, we will have found a satisfying $m$ for the RHS and we will be done. Let's try:
$$
m^k \leq \frac{(\epsilon m)^{k+1}}{(k+1)!} = \frac{\epsilon^{k+1}}{(k+1)!} m^{k+1}\Leftrightarrow \\
1 \leq \frac{\epsilon^{k+1}}{(k+1)!} m \Leftrightarrow \\
\frac{(k+1)!}{\epsilon^{k+1}} \leq m
$$
By our definition we have $m = \log(n)$, therefore we have the explicit (and awfully loose) bound:
$$
n \geq \exp\left(\frac{(k+1)!}{\epsilon^{k+1}}\right) \implies \log^k(n) \leq n^\epsilon
$$
(Note that I assume for simplicity that $c=1$. But it really doesn't matter. This shows that this proof can hold for any $c$, which means that we have the stronger result: $\log^k(n) = o(n^\epsilon)$)
A: 
Propositions 1. Function $f(x)=\frac{x^{\varepsilon}}{\ln^k{x}}$ is ascending for large $x>0$.

Because
$$f'(x)=\frac{x^{\varepsilon-1} (\varepsilon \ln^k{x}-k)}{\ln^{k+1}{x}}>0 \iff 
\varepsilon \ln^k{x}-k>0 \iff 
x> e^{\sqrt[k]{\frac{k}{\varepsilon}}}$$

Propositions 2. $\lim\limits_{x\rightarrow \infty}f(x) \rightarrow \infty$

If we assume it is bounded by a large $\alpha>0, \forall x>1$ and we know $\ln{x}$ is ascending
$$\frac{x^{\varepsilon}}{\ln^k{x}} < \alpha \iff 
1<x^{\varepsilon}< \alpha \ln^k{x} \iff 
\color{red}{0}<\varepsilon<\frac{\ln{\alpha}}{\ln{x}}+\frac{k\ln{\ln{x}}}{\ln{x}}\rightarrow \color{red}{0}, x\rightarrow\infty$$
which is a contradiction of $\varepsilon>0$. So, $f(x)$ is increasing and has no upper bound.
As a result: 
$$\lim\limits_{n\rightarrow\infty}\frac{n^{\varepsilon}}{\ln^k{n}}\rightarrow\infty \iff
\lim\limits_{n\rightarrow\infty}\frac{\ln^k{n}}{n^{\varepsilon}}=0 \iff
\ln^k{n}=o\left(n^{\varepsilon}\right)$$
