$\text{ Let }{f}_p\left(\alpha \right)=e^{\frac{i\alpha }{p^2}}\cdot e^{\frac{2i\alpha }{p^2}}\dots \dots .e^{\frac{i\alpha }{p}}, p\in N \left(\text{where }i=\sqrt{-1}\right) \text{, then find the value of }\left|\underset{n\to \mathrm{\infty }}{lim} f_n\left(\pi \right)\right|$

$\begin{array}{l}{f}_p\left(\alpha \right)=e^{\frac{i\alpha }{p^2}(1+2+\dots +p)}=e^{\frac{i\alpha }{2}\left(1+\frac{1}{p}\right)}\\ \therefore \underset{n\to \mathrm{\infty }}{lim} f_n\left(\pi \right)=\underset{n\to \mathrm{\infty }}{lim} e^{\frac{i\pi }{2}\left(1+\frac{1}{n}\right)}=e^{\frac{i\pi }{2}}\\ \Rightarrow \left|\underset{n\to \mathrm{\infty }}{lim} f_n\left(\pi \right)\right|=\left|e^{\frac{i\pi }{2}}\right|=1\\ \end{array}$