Solution
= Solution
Take the unit circle counterclockwise and $n\ge0$. The integrand is a Laurent <polynomial> with its only possible singularity at zero. Its expansion is
$$
\frac1z(z-z^{-1})^{2n}
=\sum_{j=0}^{2n}(-1)^j\binom{2n}{j}z^{2n-2j-1}.
$$
The $z^{-1}$ term is the term $j=n$, so the <residue> is $(-1)^n\binom{2n}{n}$. The <residue theorem> gives
$$
\boxed{\oint_{|z|=1}(z-z^{-1})^{2n}\frac{dz}{z}
=2\pi i(-1)^n\frac{(2n)!}{(n!)^2}.}
$$