My book states that a 3$\times$3 unitary matrix has 9 degrees of freedom, but I have trouble seeing why that is true. So I want to try and show that an $n\times n$ matrix has $n^2$ degrees of freedom.
Base case $n=1$: This is just a matrix with one entry which has length 1. It is clear that any entry with length 1 will do. So I have $1 = 1^2$ degree of freedom
$n \implies n+1$: Suppose we have an $n \times n$ unitary matrix and it has $n^2$ degrees of freedom. I want to now show that a $(n+1) \times (n+1)$ unitary matrix has $(n+1)^2$ degrees of freedom. If I add a $(n+1)$th row and column to the $n \times n$ unitary matrix, I have added $(n+1) + n = 2n + 1$ entries. For every entry $a_{ij}$ I have added, I need to ensure that each $a_{ij}^2 = 0$ in order for the matrix to be unitary. Since there are numerous ways to choose $a_{ij}$ such that $a_{ij}^2$ = 0, I have added $2n+1$ degrees of freedom. So the total degrees of freedom is $n^2 + 2n + 1 = (n+1)^2$.
Can I get some feedback on this proof? Thanks in advance!
Note It seems like my approach above is at a dead end.