Proving $n^\epsilon > \log(n)$ for sufficiently large $n$ I'd like to show that for all $\epsilon > 0$, there exists an $N$ such that for all $n \geq N$ the following holds: $n^\epsilon > \log(n)$. I'm having trouble justifying this. 
My intuition says this should hold. Writing $n = 2^k$ gives an inequality of the form $(2^\epsilon)^k > k$. This should hold for sufficiently large $k$ (and therefore $n$) because $2^\epsilon > 1$, so the function $(2^\epsilon)^k$ grows much faster than $k$. (This is also the approach described in this question.) 
However, I was wondering if there is a more "rigorous" way to justify this inequality. 
 A: If you are willing to define the logarithm of $n$ by $$\int_1^n \frac 1t \, dt$$ this follows from the fact that $t^\epsilon \to \infty$ as $t \to \infty$.
There is an index $N_1$ with the property that $t \ge N_1$ implies $t^\epsilon > \dfrac 2 \epsilon$. Thus if $n \ge N_1$ you get $$\log n - \log(N_1) = \int_{N_1}^{n} \frac 1t \, dt < \frac \epsilon 2\int_{N_1}^n t^{\epsilon - 1} \, dt < \frac \epsilon  2\int_0^n t^{\epsilon - 1} = \frac 12 n^\epsilon.$$
There is also an index $N_2$ with the property that $t \ge N_2$ implies $\log(N_1) < \dfrac 12 t^\epsilon$. Thus if $N = \max \{N_1,N_2\}$ then $$n \ge N \implies \log n < n^\epsilon.$$
A: Provided you have proved L'Hôpital's rule (as you should have in a real analysis course), this is pretty straightforward.
Fix $\epsilon>0$, we have
$$
\lim_{n\rightarrow \infty}\frac{\log n}{n^\epsilon}\stackrel{\text{L'Hôpital's rule}}{=} \lim_{n\rightarrow \infty}\frac{\frac{1}{n}}{\epsilon n^{\epsilon-1}}=0<1
$$
Proving that for sufficiently large $n$, $\log n<n^\epsilon$
A: This is a consequence of the inequality $\log x \leq x - 1$ for all positive $x$ with equality holding only when $x=1$. Given any $\epsilon>0$ choose a $\delta$ such that $0<\delta<\epsilon$. Then we have $$\delta\log x=\log x^{\delta} <x^{\delta} - 1$$ or $$\log x<\frac{x^{\delta}} {\delta} $$ and this is clearly less than $x^{\epsilon} $ if $x^{\epsilon-\delta} >1/\delta$. This happens for all large values of $x$ because $\epsilon>\delta$. 
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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\begin{align}
n^{\epsilon} = \expo{\ln\pars{n^{\epsilon}}} > 1 + \ln\pars{n^{\epsilon}} =
\ln\pars{n} + \ln\pars{\mrm{e}n^{\epsilon - 1}} > \ln\pars{n}
\quad\mbox{whenever}\quad\mrm{e}n^{\epsilon - 1} > 1
\end{align}

$\ds{\mrm{e}n^{\epsilon - 1} > 1 \implies n^{\epsilon - 1} > {1 \over \mrm{e}}\implies \bbx{\ds{n > {1 \over \mrm{e}^{1/\pars{\epsilon - 1}}}}}}$.
  The "$\ds{\epsilon = 1}$ case" is trivial.

A: We will use that if $z>0$ then $$e^z>\frac{z^2}2.\tag1$$ This follows from the power series for $e^z.$
Given $\epsilon>0.$ If $$n>e^{2/\epsilon^2}\tag2$$ you get $$\begin{align}n^{\epsilon}&=e^{\epsilon\log n}\\&> \frac{\epsilon^2\log^2n}{2}&\text{by }(1)\\&=\frac{\epsilon^2\log n}{2}\log n\\&>\log n&\text{by }(2)\end{align}$$

We can do better than $(1).$ We are essentially using, for $z>0$:
$$\frac{e^z-1}{z}\geq \frac{z}{2}.$$
That is a weak inequality. We can do much better, using the mean value theorem:
$$\begin{align}\frac{e^z-1}{z}&=\frac{e^{z}-e^{z/2}}{z}+\frac{e^{z/2}-e^{0}}{z}\\
&=\frac12\left(\frac{e^{z}-e^{z/2}}{z/2}+\frac{e^{z/2}-e^{0}}{z/2}\right)\\&\geq \frac{1}{2}e^{z/2}&\text{MVT applied to each term}
\end{align}$$
So $$e^{z}> \frac{1}{2}e^{z/2}z\tag3$$
If $e^{z/2}>\frac{2}{\epsilon}$ then $e^z\geq \frac{4}{\epsilon^2}.$
So when $z=\epsilon \log n$ you need $$n>\left(\frac{2}{\epsilon}\right)^{2/\epsilon}.$$
Then: $$\begin{align}n^{\epsilon}&=e^{\epsilon\log n}\\
&\geq \frac{1}{2}e^{\epsilon/2 \log n}(\epsilon\log n)&\text{by }(3)\\&\geq \frac{1}{2}\cdot \frac{2}{\epsilon}\cdot\epsilon\log n&\text{since }\epsilon/2\log n>\log(2/\epsilon)\\&=\log n
\end{align}$$
You can probably do better for lower bounds of $e^{z}/z.$

If $\epsilon=\frac1{10},$ the first approach gives about $n>7\cdot 10^{86}.$
The second approach gives $n>10^{26}.$
The actual lower bound is $n>4\cdot 10^{15}.$
For $\epsilon>0$ small, solving:
$$e^{\epsilon y}=y$$ is the same $-\epsilon ye^{-\epsilon y}=-\epsilon.$
or $$y=-\frac{1}\epsilon W_1(-\epsilon)$$ where $W_1$ is a branch of the Lambert $W$-function.
If $$n\geq \exp\left(-\frac1{\epsilon}W_1(-\epsilon)\right)$$
$W_1(-\epsilon)$ is only defined for $\epsilon\leq\frac1e.$
WolframAlpha can compute $W_1(x)$ as ProductLog(-1,x).
That said, $W_1(-\epsilon)$ only exists because $\lim_{x\to\infty} xe^{-x}\to 0,$ which is really what we are proving. So we can’t use $W_1$ directly to prove this theorem. Indeed, we can see the first to proofs as giving proof there is a value for $W_1(x)$ for any $-1/e\geq x>0.$

One last proof.
Let $f(x)=\frac{e^{x}}{x}.$ Then $$f’(x)=f(x)\cdot\left(1-\frac{1}x\right).$$
Use this to show that $f(x)$ is unbounded, and then use it to prove our result.
