= Solution
Set $p_*(x)=2^{1-n}T_n(x)$, a monic <Chebyshev polynomial>. Its norm on $[-1,1]$ is $2^{1-n}$. At the ordered points $x_j=\cos((n-j)\pi/n)$, its values alternate between the two extreme values $\pm2^{1-n}$.
Approximate $f(x)=x^n$ by <polynomials> of degree at most $n-1$. The error corresponding to $q_*=x^n-p_*$ is exactly $p_*$. The <Chebyshev alternation theorem> therefore makes $q_*$ the unique best approximant. For any monic $p$, the <polynomial> $q=x^n-p$ is an admissible approximant, so
$$
\boxed{\|p\|_\infty=\|x^n-q\|_\infty\geq\|p_*\|_\infty=2^{1-n}.}
$$
Equality holds precisely for $p=p_*$. Alternatively, a smaller norm would force $p-p_*$ to change sign between each consecutive pair of alternating extrema, giving at least $n$ zeros to a <polynomial> of degree at most $n-1$.
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