Every full rank matrix has a singular value decomposition of the form
$$
\mathbf{A} = \mathbf{U} \, \mathbf{S} \, \mathbf{V}^{*}
$$
where $\mathbf{S}$ is a diagonal matrix and the domain matrices are unitary:
$$
\mathbf{U}\,\mathbf{U}^{*} = \mathbf{U}^{*}\mathbf{U} = \mathbf{I}, \quad
\mathbf{V}\,\mathbf{V}^{*} = \mathbf{V}^{*}\mathbf{V} = \mathbf{I}.
$$
The Moore-Penrose pseudoinverse matrix is given by
$$
\mathbf{A}^{\dagger} = \mathbf{V} \, \mathbf{S}^{-1} \, \mathbf{U}^{*}.
$$
Show that for the full rank matrix the pseudoinverse $\mathbf{A}^{\dagger}$ is equivalent to the classic inverse $\mathbf{A}^{-1}$.
We define the classic matrix as having the property that
$$
\mathbf{A}^{-1} \mathbf{A} = \mathbf{A}\,\mathbf{A}^{-1} = \mathbf{I}.
$$
Show that the pseudoinverse has this property.
$$
\begin{align}
\mathbf{A}^{-1} \mathbf{A}
&= \left( \mathbf{V} \, \mathbf{S}^{-1} \, \mathbf{U}^{*} \right) \left( \mathbf{U} \, \mathbf{S} \, \mathbf{V}^{*} \right) \\
&= \mathbf{V} \, \mathbf{S}^{-1} \underbrace{\left( \mathbf{U}^{*}\mathbf{U}\right)}_{\mathbf{I}} \mathbf{S} \, \mathbf{V}^{*} \\
&= \mathbf{V}\underbrace{\mathbf{S}^{-1}\mathbf{S}}_{\mathbf{I}}\mathbf{V}^{*} \\
&= \mathbf{V}\,\mathbf{V}^{*} \\
&= \mathbf{I}
\end{align}
$$
Showing $\mathbf{A}\,\mathbf{A}^{\dagger} = \mathbf{I}$ uses the similar machinations.
The presumption of full rank eliminates the nullspaces and provides a full rank, diagonal matrix for $\mathbf{S}$.