Using Binomial Theorem to prove the following $$\large\sum_{j=0}^n (-1)^j {n\choose j}={n\choose 0}-{n\choose 1}+.....+\pm{n\choose n}=0 $$
I'm confused by the last part of the equation $\pm$. it seems imply that the sum would be equal to 0 no matter the $n$ is even or odd ? 

HINT
$$(1-1)^n =  \large\sum_{j=0}^n  \binom{n}{j}(1)^{n-j}(-1)^{j} $$
