= Solution
Put $x=\cos\theta$. The trigonometric identity
$$
\cos((n+1)\theta)+\cos((n-1)\theta)=2\cos\theta\cos(n\theta)
$$
gives
$$
\boxed{T_{n+1}(x)=2xT_n(x)-T_{n-1}(x).}
$$
Starting with $T_0=1$ and $T_1=x$, this recurrence proves inductively that each <Chebyshev polynomial> is a <polynomial>. For $n\geq1$, the highest-degree term in $2xT_n$ cannot cancel against $T_{n-1}$, so the degree increases by one and its <leading coefficient> doubles. Thus
$$
\boxed{\deg T_n=n,\qquad \operatorname{lc}(T_n)=2^{n-1}\quad(n\geq1).}
$$
The constant <polynomial> $T_0$ has <leading coefficient> one.
For $n\geq1$, the points $x_j=\cos(j\pi/n)$, $j=0,\ldots,n$, give $T_n(x_j)=(-1)^j$. They are all the points where $|T_n|=1$, since $|\cos(n\theta)|=1$ exactly when $n\theta$ is an integer multiple of $\pi$. Therefore there are \b[$n+1$ equioscillation points], including the two endpoints. Reversing their order makes an increasing alternating sequence. For $n=0$ the <polynomial> is constant: every point has unit value, but an alternating sequence has at most one point.
Now take $n\geq1$ and $c=2^{1-n}$. Since $cT_n$ is monic, $p_0(x)=x^n-cT_n(x)$ has degree at most $n-1$, and $\|x^n-p_0\|_\infty=c$. This proves the upper bound for the best error.
For the lower bound, suppose a <polynomial> $p$ of degree at most $n-1$ had $\|x^n-p\|_\infty<c$. At each of the $n+1$ ordered extrema, $h=p-p_0$ must have the same strictly nonzero sign as $T_n$: if $cT_n=c$, then $|c-h|<c$ gives $h>0$, and if $cT_n=-c$, then $|-c-h|<c$ gives $h<0$. Between every consecutive pair, the <intermediate value theorem> gives a zero of $h$. These are $n$ distinct zeros, impossible for a nonzero <polynomial> of degree at most $n-1$; $h=0$ is incompatible with the strict error improvement. This first-principles sign argument proves the <monic Chebyshev extremal polynomial> bound without using the alternation theorem:
$$
\boxed{E_{n-1}(x^n)=2^{1-n}.}
$$
The displayed $p_0$ supplies an actual best approximating <polynomial>.
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