Prove or disprove: $n^{\log_2 n} = \mathcal{O}(a^n)$ for all $a > 1$

Question

I am not sure how to solve this problem:

Prove or disprove: $n^{\log_2 n} = \mathcal{O}(a^n)$ for all $a > 1$

I have tried calculating $\lim_{n\to\infty} \frac{n^{\log_2 n}}{a^n}$ to see if it equals zero (which would imply that this equals $\mathcal{O}(a^n)$), but I don't think this path would work.