I was reading the following question, A function is $L^2$-differentiable if and only if $\xi\widehat{f}(\xi) \in L^2$., and in the accepted answer there is a bound $$\left\lvert\frac{e^{2\pi ih\xi}-1}{h} \right\rvert \leqslant 2\pi \lvert \xi\rvert$$ and although it may not be complicated i got stuck on showing this bound, i do see that
$$ \lim_{h \to 0} \left\lvert\frac{e^{2\pi ih\xi}-1}{h} \right\rvert = 2\pi \lvert \xi\rvert$$
So i thought of showing this function is an increasing function but all i got was a mess on calculations, is there an easier way to see that bound?
Edit: i dont know what my idea was but definetely increasing function is not a way to show the bound
$|e^{2\pi i\xi}-1|=2\pi |\xi||\int_0^{h} e^{2\pi i t\xi} dt|\leq 2\pi |\xi||\int_0^{h} 1 dt| \leq 2\pi |\xi|$ if $|h| \leq 1$.