Matrices homework problem Any help with this question would be greatly appreciated!
I understand I should be using invertibility and skew-symmetric so ,
Inverse: $M.M = I$ and,
Skew-symmetric: $m_{ij}$ = Transpose of $m_{ji}$ and,
Transpose of $M = -M$
but I have played around with this and can't quite figure out how I can prove that $n$ is even.

Let $M$ be an $n × n$ matrix.
  $M$ is called skew-symmetric if its transpose is equal
  to $−M$. Assume that $M$ is skew-symmetric and invertible. Prove
  that $n$ is even.

 A: 
$A=(a_{ij})_{n\times n}$ is invertable, if $\det A \neq 0$
$A=(a_{ij})_{n\times n}$ is a skew-symmetric matrix, so $A^T=-A$ i.e., if $a_{ij}=-a_{ji}$

${}$

For a $n \times n$ matrix $A$,
$1.$ $\det (A) = \det (A^T)$ and
$2.$ $\det (cA)=c^n \det A$

${}$
So here $\det (M)=\det (M^T)=\det (-M)=(-1)^n \det (M)$, as $M=(m_{ij})_{n\times n}$ is a Skew-symmetric matrix.
Now if $n$ is odd, then $\det (M) =-\det (M)\implies 2\det (M)=0\implies \det (M)=0$, i.e., $M$ is not invertible.

The determinant of an $n×n$ skew-symmetric matrix $A$ is zero if $n$ is odd.
${}$



If the dimension of a skew-symmetric matrix is even i.e., if $n$ is even, then $$\det (M)=\det (M^T)=\det (-M)=(-1)^n \det (M)=\det (M)$$ so we do not have any conclusion about $\det(M)$.
The determinant of a skew-symmetric matrix $M$ of even dimension is the square of a polynomial, called the Pfaffian, in the entries of $M$.
As a corollary, this determinant is thus non-negative.
For example: $M=\begin{pmatrix} 
0 & k \\
-k & 0 
\end{pmatrix}$ be a skew-symmetric matrix.
Now $\det(M)=\begin{vmatrix} 
0 & k \\
-k & 0 
\end{vmatrix}=0-k(-k)=k^2$
so the Pfaffian is $k$.
In the $2×2$ case, then, any skew-symmetric matrix is nonsingular with a positive determinant, except the zero matrix.
Ref:
$1.$ https://www.quora.com/What-is-the-determinant-of-every-skew-symmetric-matrix
$2.$ http://mathworld.wolfram.com/Pfaffian.html
