Why $\overline{\hat{f}(k)}=\hat{f}(-k)$ If $f \in \mathbb{R}$

Question

Proposition: Suppose $f$ real valued $2\pi$ periodic and integrable function on a closed interval of $2\pi$ length. Then $$ \overline{\hat{f}(k)}=\hat{f}(-k)$$ Proof:
We can easily see that: $$ \overline{\hat{f}(k)}= \overline{\frac{1}{2\pi}\int_{-\pi}^{\pi}f(x)e^{-ikx}dx} =\frac{1}{2\pi}\int_{-\pi}^{\pi}\overline{f(x)e^{-ikx}}dx=\frac{1}{2\pi}\int_{-\pi}^{\pi}f(x)e^{ikx}dx=\frac{1}{2\pi}\int_{-\pi}^{\pi}f(x)e^{-i(-k)x}dx=\hat{f}(-k)$$ I can't see we this is true only if $f$ is real valued fuction. I know that $f$ is real valued fuction if $f=\overline{f}$, but where in the proof we use this fact ?