Is $(-1)^n$ equal to $(-1)^{n+2}$ for any natural number n? I just got introduced to proofs by induction and, after doing the whole procedure of algebraic manipulations I got everything I wanted to have in the expression except for a $(-1)^{n+2}$, in which I ended up with a $(-1)^n$ instead (n being a natural number). Considering that $(-1)^n$ is 1 if n is even and -1 if n is odd alongside the fact that the parity of n is the same as n+2 for every natural number, can I safely say that $(-1)^n$ = $(-1)^{n+2}$?
 A: You can just write further as:
$$(-1)^n=(-1)^n\cdot 1=(-1)^n(-1)^2=(-1)^{n+2}$$
where you can add/remove steps depending upon the background.
Hope this helps. Ask anything if not clear :)
A: The arguments that you made do hold, but since you are working with proof by inductions, you could use proof by inductions to prove this statement, too.
The proof by induction would look something like this:
Base case: n=0
$(-1)^0 = (-1)^{0+2}$
$1 = (-1)^2$
$1=1$
Inductive step: Assume that it holds for k=n. Show then that it works for k+1
$(-1)^k = (-1)^{k+2}$
$(-1)^k \cdot (-1) = (-1)^{k+2} \cdot (-1) $
$(-1)^{k+1} = (-1)^{(k+1)+2}$
End of proof.
A: We can use D'moivre identity to prove $(-1)^n = (-1)^{n + 2}$.
D'moivre idenity state that for real number $\alpha$ we have
$$(\cos \alpha + i\sin \alpha)^n = (\cos n\alpha + i\sin \alpha n)$$
for all natural number $n$.
Since $-1 = \cos\pi + i\sin \pi$ we have
\begin{align*}
(-1)^n &= (\cos \pi + i\sin \pi)^n\\
&=(\cos\pi n + i\sin \pi n) \\
&= (\cos (\pi n + 2\pi) + i\sin (\pi n + 2\pi)) \\
&=\cos(\pi(n + 2)) + i\sin (\pi(n+2))\\
&= (\cos \pi + i\sin \pi)^{n + 2}\\
&= (-1)^{n + 2}
\end{align*}
