Prove that the determinant of $ A^{-1} = \frac{1}{det(A)} $- Linear Algebra

If I have a single matrix A that is non-singular, how can I prove the determinant of its inverse = $\frac{1}{\det(A)}$?

Prove: $$ \det(\mathbf{A^{-1}}) = \frac{1}{\mathbf{\det(A)}} $$

I know that $(A)(A^{-1}) = I$, but I am not sure what to do with that knowledge.

first of all we know that

$$\det(A \cdot B)=\det(A)\times\det(B)$$

also we know that

$$A\times A^{-1}=I$$

we know that

$$\det(A \cdot A^{-1})=\det(I)$$

or

$$\det(A)\times\det(A^{-1})=\det(I)$$

Can you continue from this? Ask yourself: what is $\det(I)$?) Take the example of the $3\times 3$ identity matrix:

$$ I = \begin{pmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}$$

$$ \det(I) = 1$$

So,

$$\det(A)\times\det(A^{-1})=1$$

or

$$\boxed{\det(A^{-1})=\frac{1}{\det(A)}}$$