Answer
The statement is correct.
Work Step by Step
Since the matrix $A$ is a square matrix of order $n$ and a skew-symmetric matrix, then $A^{T}=-A$ .
Let the general form of the matrix $A$ be as follows
$A=\left[\begin{array}{ccccc}
a_{11} & a_{12} & & & & a_{1n} \\
a_{21} & a_{22} & & .. & & a_{2n} \\
& . & . & . & &. \\
& & && . & . \\
a_{n1} & a_{n2} & . &&& a_{nn} \\
\end{array}\right]$
Since $|A^{T}|=|A|$, using the assumption $A^{T}=-A$, we obtain $|-A|=|A|$ and then we have
$|A|= |-A|=\left|\begin{array}{ccccc}
-a_{11} & -a_{12} & & & & -a_{1n} \\
-a_{21} & -a_{22} & & .. & & -a_{2n} \\
& . & . & . & &. \\
& & && . & . \\
-a_{n1} & -a_{n2} & . &&& -a_{nn} \\
\end{array}\right|$
By taking the common factor $(-1)$ of $n$ rows of the determinant $|-A|$, we have
$|A|=|-A|=(-1)(-1)(-1) (-1) * \left|\begin{array}{ccccc}
a_{11} & a_{12} & & & & a_{1n} \\
a_{21} & a_{22} & & .. & & a_{2n} \\
& . & . & . & &. \\
& & && . & . \\
a_{n1} & a_{n2} & . &&& a_{nn} \\
\end{array}\right|=(-1)^{n}|A|$
Thus $|A|=(-1)^{n}|A|$
