Let $f: \mathbb{C} \to \mathbb{C} $ be a holomorphic function with $$ \lim_{\lvert z \rvert\to\infty} \frac{f(z)}{z^{n-1}} = 0$$ for some $n\in\mathbb{N}$.
How can I prove that $$ \lim_{r\to\infty} \int_\gamma f(z) \exp (iz^n) dz = 0$$ with $\gamma: [0, \frac{2\pi}{n}] \to \mathbb {C} , \, \, \, \, \gamma(t) = re^{it} $.
My attempt is that $f$ has to be a polynom of grade $(n-2)$ with coefficients $$a_k = \frac{1}{2\pi i} \int_{\lvert z \rvert = r} \frac{f(z)}{z^{k+1}}dz$$ so for $k=n-2$ I receive $$a_{n-2} = \frac{1}{2\pi i} \int_{\lvert z \rvert = r} \frac{f(z)}{z^{n-1}}dz$$ which looks rather similar to my first assumption.
Can anyone give me a clever hint how to proceed ?