= Solution
Since $M^T=-M$, $(M^2)^T=(M^T)^2=M^2$, so $M^2$ is a real <symmetric matrix>. Moreover,
$$
\boxed{x^TM^2x=-x^TM^TMx=-\|Mx\|^2\leq0,}
$$
proving that it is a <negative semidefinite matrix>.
For odd $n$, <determinant> invariance under transpose gives
$$
\det M=\det M^T=\det(-M)=(-1)^n\det M=-\det M,
$$
so $\det M=0$. Hence $M$ and $M^2$ have a zero <eigenvalue>. Finally, if $M^2w=0$, then $0=w^TM^2w=-\|Mw\|^2$, so $Mw=0$. Thus
$$
\boxed{\ker M^2=\ker M,}
$$
where equality of these <kernels> includes the reverse implication, which is immediate.
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