Understanding Inverse Matrices with Powers

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If [tex]A = [a_{ij}]^{n\times n}[/tex] is invertible, show that [tex](A^{2})^{-1} = (A^{-1})^{2}[/tex] and [tex](A^{3})^{-1} = (A^{-1})^{3}[/tex]

So basicaly we have a square matrix with elements [tex]a_{ij}[/tex]. This looks slightly familar to [tex](A^{T})^{-1} = (A^{-1})^{T}[/tex]. Are [tex]A^{2}[/tex] and [tex]A^{3}[/tex] meant to be the elements of the matrix raised to those respective powers? Or does it mean that the matrix is [tex]2\times 2[/tex] or [tex]3\times 3[/tex]?
 
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The matrix is nxn. A^2 means AA.

As 90% of linear algebra proofs, these problems are solvable in 2-3 lines. If you really don't find it, I can start you and you will find it imidiately. Laying the first equation is the hardest.
 
thanks. I got it, just wasn't clear about the notation.