maximum of monic polynomial on unit circle is 1 implies $p(z)=z^n$

Let $p(z)=z^n+a_{n-1}z^{n-1}+\cdots+a_0$.

Then, $\forall z\neq 0:p(\frac{1}{z})=z^{-n}+a_{n-1}z^{-n+1}+\cdots+a_0$.

Let $q(z)=z^n\cdot p(\frac{1}{z})=1+a_{n-1}z+\cdots+a_0z^n$.

Observe that $$\max\limits_{|z|=1}|q(z)|=\max\limits_{|z|=1}|z^n\cdot p(1/z)|=\max\limits_{|z|=1}|p(1/z)|=\max\limits_{|z|=1}|p(z)|$$

The last step is due to $\{1/z\mid z\in\mathbb{C}:|z|=1\}=\{z\mid z\in\mathbb{C}:|z|=1\}$.

Now, $q(0)=1$, so $|q(0)|=1$ and, by the maximum principle, $\max\limits_{|z|=1}|q(z)|\geq1$.

Finally, $\max\limits_{|z|=1}|p(z)|\geq1$.

Now, if $p(z)=z^n$ then clearly $\max\limits_{|z|=1}|p(z)|=1$.

Suppose that $\max\limits_{|z|=1}|p(z)|=1$, then also $\max\limits_{|z|=1}|q(z)|=1$.

Again, by the maximum principle, we have that $q\equiv 1$.

Therefore, $\forall z\neq 0:p(1/z)=1/z^n$, so $\forall z\neq 0:p(z)=z^n$.

Finally from continuity $p(0)=0$ and we get that $\forall z\in\mathbb{C}:p(z)=z^n$.

You've shown that the mean value of $|p(z)|$ on the unit circle is at least $1$, so to show that the maximum is greater than $1$, you just have to do is show that $|p(z)|$ is not constant.

Suppose for the sake of contradiction that $|p(z)|$ is constant on the unit circle. Then $iz p'(z)$, the directional derivative of $p(z)$ in a direction tangent to the unit circle at $z$, must be an imaginary multiple of $p(z)$ everywhere on the unit circle; that is, the (necessarily meromorphic!) function $q(z) = \frac{p(z)}{z p'(z)}$ must be real wherever $|z| = 1$. By a corollary of the maximum modulus principle, $q(z)$ is equal everywhere to some constant $c$, and we can write $p(z) = cz p'(z)$. By comparing leading coefficients of $p(z)$ and $zp'(z)$ we get that $c$ has to equal $1/n$, but in that case, none of the nonzero trailing coefficients of $p(z)$ and $czp'(z)$ match up, so $p(z) = \frac{1}{n} z p'(z)$ can only hold if $p(z) = z^n$.