Proof that $(-1)^{n+1}$ = $(-1)^{(n-1)}$ I have been working on proving that $(-1)^{n+1}$ = $(-1)^{(n-1)}$ by mathematical induction.  Is this the correct proof?
Formula:
$(-1)^{n+1} = (-1)^{n-1}$
Chose $n = 2$.  $-1 = -1$.  True for $n = 2$.
Assumption:
$(-1)^{n+1} = (-1)^{n-1}$
Case n+1:
$(-1)^{(n+1)+1} = (-1)^{(n+1)-1}$
$(-1)^{n+2} = (-1)^{n}$
Divide by $-1$ on both sides.
$(-1)^{n+1} = (-1)^{n - 1}$
This is the same as the assumed formula, so the statement is proven.
 A: Your proof is not quite correct. 
First, you made a typo, it is $(-1)^{n+1}$ and not $(-1)^n$ at the beginning.
Then, you can not start from $(-1)^{(n+1)+1}=(-1)^{(n+1)-1}$ because you don't know at first this is true. 
Instead, you have to do the reasoning in the other direction, starting from what you know:
$$(-1)^{n+1}=(-1)^{n-1}$$
and multiply by $(-1)$ to get the result for the case $n+1$.
A: Your assumption really should be:
Assume $(-1)^{n+1}=(-1)^{n-1}$ for all numbers up to and including $n$. 
Then when you finish your induction proof, you must realize that have really proven it only for integers 2 and above (as your base case is 2--you could have chosen 0 or any other integer for your base, nothing special about 2 here). You could make your base case lower, but to prove this using induction for all integers, you have to prove things in the other direction, for number flowing off in the direction below your base case (1,0,-1,-2,-3,...), perhaps with a separate induction.
It's much more complicated than the simple way of showing this identity true for all integers in the comments above.
