Proving $n^\epsilon > \log(n)$ for sufficiently large $n$

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

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 $n$, $\log n<n^\epsilon$