Relation between fourier coefficients of $f\in \mathcal{C}^1[-\pi, \pi]$ and $f'$

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I'm given $f\in \mathcal{C}^1[-\pi, \pi]$ with $f(-\pi)=f(\pi)$. It's fourier coefficients are given by:

$$\gamma_n=\frac{1}{2\pi}\int_{-\pi}^{\pi}e^{-int}f(t)dt,\ n\in \mathbb{Z}$$

And now I'm asked to express the fourier coefficients of $f'$ in terms of the $\gamma_n$.

So my reasoning is that we have $f=\sum_{-\infty}^{\infty}\gamma_n e^{int}$, and so we just find that $f'=(\sum_{-\infty}^{\infty}\gamma_n e^{int})'=\sum_{-\infty}^{\infty}in\gamma_n e^{int}$. So that the coefficients of $f'$ are just $in\gamma_n$.

However I don't think this is correct. Because as far as I can tell I did not use the fact that $f(-\pi)=f(\pi)$, and it's probably not given for no reason.

So can someone tell me what I'm doing wrong here?

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The Fourier coefficients of $f'$ are given by

$$c_n = \frac{1}{2\pi}\int_{-\pi}^\pi e^{-int}f'(t)\, dt\quad (n\in \Bbb Z).$$

By integration by parts,

$$c_n = \frac{e^{-int}f(t)}{2\pi}\bigg|_{-\pi}^\pi - \frac{1}{2\pi}\int_{-\pi}^\pi (-in e^{-int})f(t)\, dt = in\gamma_n,$$

using the boundary condition $f(-\pi) = f(\pi)$.