How to prove $(f \circ\ g) ^{-1} = g^{-1} \circ\ f^{-1}$? (inverse of composition)

I'm doing exercise on discrete mathematics and I'm stuck with question:

If $f:Y\to Z$ is an invertible function, and $g:X\to Y$ is an invertible function, then the inverse of the composition $(f \circ\ g)$ is given by $(f \circ\ g) ^{-1} = g^{-1} \circ\ f^{-1}$.

I've no idea how to prove this, please help me by give me some reference or hint to its solution.

You put your socks first and then your shoes but you take off your shoes before taking off your socks.

Use the definition of an inverse and associativity of composition to show that the right hand side is the inverse of $(f \circ g)$.

$$\begin{align} & \text{id} \\ =& f \circ f^\circ \\ =& f \circ \text{id} \circ f^\circ \\ =& f \circ (g \circ g^\circ) \circ f^\circ \\ =& f \circ g \circ g^\circ \circ f^\circ \\ =& (f \circ g) \circ (g^\circ \circ f^\circ) \end{align}$$

Therefore $(f \circ g)^\circ = g^\circ \circ f^\circ$.