Is $(1-p)^0$ equal to $1$ or $-1$? Context:

I would like to proof that $C^2_0 p^0 (1-p)^2 + C^2_1 p^1 (1-p) + C^2_2 p^2 (1-p)^0 = 0$. When I do derivation for the equation, I stopped in the following point: 
$(1-p)^2 + 2p(1-p) + p^2$

I would like to find out the output of the first term, $(1-p)^0$, is it 1 or -1?

Ibrahim
$$x^0=1\space(\forall)\space x\in\Bbb{R}^*$$

I did say $\Bbb{R}^*$, because $0^0$ is not unanimously agreed upon or might be context-dependent.
For all  $1-p&gt;0$ it's $1$, otherwise it's not defined.