If $T\in B(H)$ for some Hilbert space $H$, is a normal operator and $T^2=I$, then $T=T^\*$. It seemed simple when I first saw the claim, but I'm having trouble showing it. I know that it implies 

$I=(T^\*)^2=(T^\*T)^2=(TT^\*)^2=(T^\*T)(TT^\*)=(TT^\*)(T^\*T)$, etc. 

Does the result come from looking at inner products, or is there a nice string of equalities to look at? Or is there a good way to show that $TT^\*=I$?