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.