Under which conditions $-(-c)^{-n}c^n=(-1)^{1-n}$?

Question

Set $c>0$ and define $0\leq n \leq m$ and integer.

Do we need certain conditions to be satisfied so we have $$-(-c)^{-n}c^n=(-1)^{1-n}?$$

When we try to simplify the LHS in Mathematica, we don't get the expression on the RHS.

Answer 1

Not sure where you're using $m$ but I conclude $$-(-c)^{-n}c^n = -\frac{1}{(-c)^n}c^n = -\big(\frac{c}{-c}\big)^n = -(-1)^n = (-1)^{1-n}.$$

This shouldn't need more conditions than $c \neq 0$ and $n \in \mathbb{N}$.

Answer 2

Given $c$ a positive number and integer $n$, we have

$$-(-c)^{-n}c^n\underset{a^{-n}=\frac 1{a^n}}=-\frac {c^n}{(-c)^n}\underset{\frac{a^n}{b^n}=(\frac ab)^n}=-\left(\frac c{-c}\right)^n\underset{\frac a{-a}=-1}=-(-1)^n\underset{a\cdot a^n=a^{n+1}}=(-1)^{n+1}$$

Note that $(-1)^{n+1}=(-1)^{1-n}(-1)^{2n}=(-1)^{1-n}((-1)^2)^n=(-1)^{1-n}1^n=(-1)^{1-n}$.

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