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. 

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 $$n \ge N \implies \log n < n^\epsilon.$$
