For which values of $N$ is $x^N \ge \ln x$ for all $0 < x < \infty$? 
Find, with proof, the smallest value of $N$ such that $$x^N \ge \ln x$$ for all $0 < x < \infty$. 

I thought of adding the natural logarithm to both sides and taking derivative. This gave me $N \ge \frac 1{\ln x}$. However, is there a better way to this?
Please note that I would like to see only a hint, not a complete solution.
If anything, I am made aware that the answer is $N \ge \frac 1e$.
 A: As you increase $n$, the graph of $x^n$ widens and goes away from $\ln x$. At the smallest value of $n$ (so that the inequality holds), the two graphs touch each other. So the two slopes at that point of touching must be same. $$\implies nx^{n-1}=\dfrac {1}{x}$$ $$\implies x^{n}=\dfrac {1}{n}=\ln x$$ as $x^n=\ln x$ at that point. Now $$x^n=\ln x\implies n\ln x=\ln (\ln x)\implies 1=\ln \left (\dfrac {1}{n}\right)$$
$$\implies n=\dfrac {1}{e}$$ Therefore $n\geq \dfrac {1}{e}$
NOTE: Only when the two curves are tangent at the point of intersection, will the inequality hold, because there are three possibilities: either cut tangentially, or cut once and go down(thus violating the inequality) or never cut and stay at the top(never satisfying the equality sign).
A: Hint for one possible approach:
Consider the function $f:(0,\infty) \to \mathbb{R}$ defined by $f(x) = \frac{\log x}{x^N}$. Take the derivative, and you find that $f$ attains a global maximum value at $x=e^{\frac{1}{N}}$.
Now solve the following equation for $N$: $$f(e^{\frac{1}{N}})=1$$ Solving this gives you $N = \frac{1}{e}$, and I'll leave it to you to think about why solving this equation gives you the minimal $N$-value that you desire.
A: first you can show that the line $ y = kx $ touches at $x = e, y = 1, k = \frac 1 e.$  that is $$kx > \ln(x) \text{ for } k > \frac1e \text{ and } k x = \ln x \text{ for } x = e, k = \frac 1e.\tag 1$$
we have $$x^N > \ln x \implies N\ln x > \ln(ln(x) \implies N > \frac 1e. $$
$$\text{ so the smallest value of } N \text{ is }\frac 1e.$$
