An $n\times n$ matrix $A$ over the field of complex numbers is said to be unitary if $AA^* = I$ ($A^*$ denotes the conjugate transpose of $A$). If $A$ is unitary, show that $|\text{det}(A)|=1$.
My attempt: Suppose $A\in M_n(\Bbb{C})$ such that $A$ is unitary. Let $\overline{A}=(\overline{A_{ij}})\in M_n(\Bbb{C})$. Then $A^*=(\overline{A})^t$ is conjugate transpose of $A$. So $(A^*)_{ij}= \overline{A_{ji}}$. Since $A$ is unitary, we have $\text{det}(AA^*)= \text{det}(A)\cdot \text{det}(A^*)= \text{det}(I)=1$. By theorem 2 section 5.3, $\text{det}(A^*)=\sum_{\sigma \in S_n}(\text{sgn}\sigma)\prod_{i=1}^nA^*(i,\sigma (i)) =\sum_{\sigma \in S_n}(\text{sgn}\sigma)\prod_{i=1}^n\overline{A(\sigma(i),i)}$. By theorem 1.31 (b) of Baby Rudin, $ \prod_{i=1}^n\overline{A(\sigma(i),i)}=\overline{\prod_{i=1}^nA(\sigma(i),i)}$. It’s easy to check, $(\text{sgn}\sigma)\cdot \overline{\prod_{i=1}^nA(\sigma(i),i)} =\overline{\text{sgn}(\sigma)\cdot \prod_{i=1}^nA(\sigma(i),i)}$. So $\text{det}(A^*)=\sum_{\sigma \in S_n} \overline{\text{sgn}(\sigma)\cdot \prod_{i=1}^nA(\sigma(i),i)}$. Since $|S_n|=n!$, we have $$\sum_{\sigma \in S_n} \overline{\text{sgn}(\sigma)\cdot \prod_{i=1}^nA(\sigma(i),i)}= \overline{\sum_{\sigma \in S_n}\text{sgn}(\sigma)\cdot \prod_{i=1}^nA(\sigma(i),i)}= \overline{\text{det}(A^t)}= \overline{\text{det}(A)}$$ by theorem 1.31 (a) of Baby Rudin. Thus $\text{det}(A^*)= \overline{\text{det}(A)}$ and $\text{det}(A)\cdot \overline{\text{det}(A)}=1$. By definition of absolute value, $|\text{det}(A)|=\left(\text{det}(A)\cdot \overline{\text{det}(A)}\right)^{\frac{1}{2}}=(1)^{\frac{1}{2}}=1$. Hence $|\text{det}(A)|=1$. Is my proof correct?
Edit for reopen: I don’t think, this post is duplicate (or similar) to Proof: |detA| = 1 if A is unitary (that is, if $AA^* = I_n).$ Does it make sense? post. Precisely, that post use induction approach and cofactor expansion of determinants. My proof is more straightforward and I used Leibniz formula for determinant. By uniqueness of determinant function, they are equal but in some certain circumstances one expression is better to use than other.