The proof is given as,

Suppose n is any odd integer. By definition of odd integer, \(\displaystyle{n}={2}{p}\ +\ {1}\) for some integer p.

\(\displaystyle{\left(-{1}\right)}^{{{n}}}={\left(-{1}\right)}^{{{2}{p}\ +\ {1}}}\)

\(\displaystyle={\left(-{1}\right)}^{{{2}{p}}}\ {\left(-{1}\right)}\)

\(\displaystyle={\left({1}\right)}^{{{p}}}\ {\left(-{1}\right)}\)

\(\displaystyle={1}\ \times\ {\left(-{1}\right)}\)

\(\displaystyle=\ -{1}\)

Therefore, if n an odd integer, then \(\displaystyle{\left(-{1}\right)}^{{{n}}}=\ -{1}\).

Conclusion:

The statement, If n is any odd integer, the \(\displaystyle{\left(-{1}\right)}^{{{n}}}=\ -{1}\) is proved.