The set is non-empty because $1\in S$, so we have a non-empty subset of $\mathbb R$ that has an upper bound then $S$ has a supremum, say $\alpha$, Claim: $\alpha^{3}=5$, try to prove the claim.
When I took intro to Linear Algebra I found it useful to restrict myself to understanding $2\times2$ matrices. Once I understood a concept for those it was much easier to grasp the same concepts for higher dimensional square matrices. Rectangular matrices are not interesting in intro.
Think of matrix $A$ as a transformation. That is, take some vector in $x\in\mathbb{R}^2$. When you do $Ax$, you are in essence transforming the vector $x$ into something else. Of course, if $A$ is the identity matrix you can think of $Ax$ as $A$ transforming $x$ into itself.
So, $A$ could be a rotation transformation, or perhaps a dilation transformation, or any sort of transformation of $x$. When we say the inverse of $A$ exists, whom we denote $A^{-1}$, in essence we are saying:
Apply the transformation $A$ to $x$ to get some new vector $y$, possibly $x$ itself as described above. But, if we have $y$ and we want $x$ we need to undo what $A$ did to $x$. This is what $A^{-1}$ does. It takes $y$ in $A^{-1}y$ and undoes what $A$ did to get you back $x$. Notice that $A^{-1}$ itself is a transformation.
If $A$ rotated $x$ by 90 degrees clockwise to get $y$, then $A^{-1}$ will rotate $y$ 90 degreescounter-clockwise to get back $x$.
When $A^{-1}$ does not exist we mean, loosely speaking, that $A$ does not behave in a way that is predictable/reversible.
This concept is the same for higher dimension matrices, only the transformations are happening in 3D for $x\in\mathbb{R}^3$ and so on.
I hope this help. This is not a rigorous answer, as it is not meant to be, but it helped my intuition back then.
Best Answer
Since $P=P^2$, if $(P)^{-1}$ exists, then $(P^2)^{-1}$ exists, so $(P^{-1})=(P^2)^{-1}=(P^{-1})^2$. In fact, it can be shown that $P=I$ is the only invertible projection matrix, so the result is obvious from this fact.