Is the inverse of a symmetric matrix also symmetric? Let $A$ be a symmetric invertible matrix, $A^T=A$, $A^{-1}A = A A^{-1} = I$ Can it be shown that $A^{-1}$ is also symmetric?

I seem to remember a proof similar to this from my linear algebra class, but it has been a long time, and I can't find it in my text book.
You can't use the thing you want to prove in the proof itself, so the above answers are missing some steps. Here is a more complete proof. Given A is nonsingular and symmetric, show that $ A^{-1} = (A^{-1})^T $:

$$ I = I^T $$

since $ AA^{-1} = I $,

$$ AA^{-1} = (AA^{-1})^T $$

since $ (AB)^T = B^TA^T $,

$$ AA^{-1} = (A^{-1})^TA^T $$

since $ AA^{-1} = A^{-1}A = I $, we rearrange the left side

$$ A^{-1}A = (A^{-1})^TA^T $$

since $ A = A^T $, we substitute the right side

$$ A^{-1}A = (A^{-1})^TA $$
$$ A^{-1}A(A^{-1}) = (A^{-1})^TA(A^{-1})$$
$$ A^{-1}I = (A^{-1})^TI $$
$$ A^{-1} = (A^{-1})^T $$

and we are done.
Inverse of $A$ can be expressed as a polynomial $p(A)$ of $A$ (from Cayley-Hamilton theorem).

So it is sufficient to prove that if $A$ is symmetric then power $A^k$ is symmetric, sum of symmetric matrices is symmetric and multiply by scalar is symmetric.