So I was just looking over an old homework problem, and my proof doesn't seem right. I got full credit for it, but it seems to be circular reasoning. Please tell me if this is actually a valid proof:

Suppose $A \in M\_{n\times n}(F)$. Then $$\det(kA) = \det(kI\_nA) = \det(kI\_n)\det(A) = k^n\det(I\_n)\det(A) = k^n(1)\det(A) = k^n\det(A).$$

I am wondering if claiming that $\det(kI\_n) = k^n\det(I\_n)$ was a valid move, since that is basically what the problem asks me to prove. Is it just a given for the identity matrix? I don't remember what I was thinking when originally doing this problem.

(Side note, how do I do a vertically aligned string of equalities in LaTex? )