If A is a skew-symmetric matrix and n is an even natural number,

If $A$ is a skew-symmetric matrix, then $A^{T}=-A$.

$\left(A^{n}\right)^{T}=\left(A^{T}\right)^{n} \quad[$ For all $n \in N]$

$\Rightarrow\left(A^{n}\right)^{T}=(-A)^{n} \quad\left[\because A^{T}=-A\right]$

$\Rightarrow\left(A^{n}\right)^{T}=(-1)^{n} A^{n}$

$\Rightarrow\left(A^{n}\right)^{T}=A^{n}$, if $n$ is even or $-A^{n}$, if $n$ is odd.

Hence, $A^{n}$ is symmetric when $n$ is an even natural number.