Showing that if $n$ is odd then $(-1)^n = -1$

Question

Greets,

In the exercises, at the end of chapter 1.4, Basic Mathematics, Serge Lang

6) Prove: If $n$ is odd, then $\quad (-1)^n = -1$

How? The working I did

$$\begin{align}( -1)^n &= ( -1 )^{2m+1}\\ &= ( ( -1)^2 )^{m+1}\\ &= ( ( -1 )( -1 ) )^{m+1}\\ &= ( 1 )^{m+1} \end{align}$$

Can it get from here to $-1$?

The book shows

Let $n = 2k + 1$. Then

$$\begin{align}(-1)^n &= (-1)^{2k+1}\\ &= (-1)^{2k}(-1)\\ &= 1 \cdot (-1)\\ &= -1 \end{align}$$

I'm having trouble with how $(-1)^{2k+1}$ became $(-1)^{2k}(-1)$.

Answer 1

$( -1)^n = ( -1 )^{2m+1} \color{red}= ( ( -1)^2 )^{m+1}$

Wrong. Note that $2(m+1)=2m+2$.

$(-1)^n = (-1)^{2k+1} \color{blue}= (-1)^{2k}(-1)$, for the blue equality, answer this question: how do you define $p^q$, where $p,q\in \Bbb Z$ and $q>0$?

Answer 2

You made an error in the 2nd equality. The relevant property of exponentiation here is: $$ a^{m+n}=a^m a^n $$