The monic Chebyshev polynomial appears perhaps a bit more naturally as a unique minimizer to the following $L^2$-problem.
Problem: find a monic polynomial $p(x)$ of degree $n$ which minimizes the weighted norm $$\|p\|^2=\int_{-1}^{1}p^2(x)\frac{dx}{\sqrt{(1-x^2)}}.$$
The proof is straightforward. First, we check the orthogonality property $$\int_{-1}^{1}T_i(x)T_k(x)\frac{dx}{\sqrt{(1-x^2)}}=\delta_{ik}\frac{\pi}{2},\quad i,k=0,1,2,...,$$ which is equivalent to the orthogonality property of the sequence $\cos kx$ in $L^2(0,\pi)$. Next, we have for an arbitrary monic polynomial of degree $n$ $$p(x)=\sum\limits_{k=1}^n a_kT_k(x),\quad a_n=2^{1-n}.$$ Therefore $$\|p(x)-2^{1-n}T_n(x)\|^2=\|p\|^2+\|2^{1-n}T_n(x)\|^2-2^{2-n}\int_{-1}^{1}\sum\limits_{k=1}^n a_kT_k(x)T_n(x)\frac{dx}{\sqrt{(1-x^2)}}= $$ $$=\|p\|^2-\|2^{1-n}T_n(x)\|^2.$$ So $\|p\|\geq \|2^{1-n}T_n(x)\|,$ and the equality is possible if and only if $p(x)=2^{1-n}T_n(x)$.