Chebyshev polynomials semigroup property $T_n \circ T_m = T_{nm}$

Question

Consider set of Chebyshev polynomials $T_n(x):\mathbb{R} \to \mathbb{R}$ given by formula $$ T_n(\cos(x)) = \cos(nx) $$ I am interested in elegant way to show that Chebyshev polynomials form a semigroup, namely proof of semigroup property: $$ T_n \circ T_m = T_{nm} $$ I have proved it using explicit form of Chebyshev polynomials, but it's rather big and unpretty

Answer 1

I prefer $$(T_n\circ T_m)(\cos x)=T_n(T_m(\cos x))=T_n(\cos mx)=\cos(nmx)=T_{nm}(\cos x).$$

Answer 2

Here is an easy proof. $T_n(x)=\cos(n\arccos(x))$. Therefore $$T_n(T_m(x))=\cos(n\arccos(\cos(m\arccos(x))))=\cos(mn(\arccos(x)))=T_{mn}(x).$$