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\begin{document}
\[x^{(n+1)(n-1)}=x^{n^2-1}  \neq x^{n^2}-1\]
\[\sum_{i=0}^{\infty} e^{i+1}\]
\[\int _0^{2\pi}e^{e^{it} - it}dt\]
\[\int _0^{2\pi}e^{ - it+e^{it}}dt\]
\[\left( \frac{-1}{p} \right) = (-1)^{\frac{p-1}2}=\ldots\]
\end{document}
    

x(n+1)(n−1)=xn2−1≠xn2−1x^{(n+1)(n-1)}=x^{n^2-1} \neq x^{n^2}-1

∑i=0∞ei+1\sum_{i=0}^{\infty} e^{i+1}

∫02πeeit−itdt\int _0^{2\pi}e^{e^{it} - it}dt

∫02πe−it+eitdt\int _0^{2\pi}e^{ - it+e^{it}}dt

(−1p)=(−1)p−12=…\left( \frac{-1}{p} \right) = (-1)^{\frac{p-1}2}=\ldots