Fourier Series- solving for -cos nπ

Question

When solving for $\cos n\pi$ it becomes $(-1)^n.$

However, if you have a negative in front, it becomes $-(-1)^n.$ Can you show how they get

$-(-1)^n = (-1)^{n+1}.$

Answer 1

One may observe that $$ a^{n+1}=a\times a^n, \quad a \in \mathbb{R}, $$ giving in particular, with $a=-1$, $$ (-1)^{n+1}=(-1)\times (-1)^n=-(-1)^n. $$