Maximum modulus for polynomials from circular mean squares

Let \(p(z)=a_0+a_1z+\cdots+a_nz^n\) and suppose \(z_0\in\mathbb C\), \(|z_0|<1\), is such that \[ |p(z)|\le |p(z_0)|\qquad(|z|\le1). \] Write \[ p(z)=b_0+b_1(z-z_0)+\cdots+b_n(z-z_0)^n. \] For \(0<r<1-|z_0|\), verify \[ \frac1{2\pi}\int_{-\pi}^{\pi}|p(z_0+re^{i\theta})|^2\,d\theta =|b_0|^2+|b_1|^2r^2+\cdots+|b_n|^2r^{2n}, \] and \[ \frac1{2\pi}\int_{-\pi}^{\pi}|p(z_0+re^{i\theta})|^2\,d\theta\le |b_0|^2. \] Conclude that a nonconstant polynomial cannot attain its maximum modulus in the open unit disk, and hence \[ \max_{|z|\le1}|p(z)|=\max_{|z|=1}|p(z)|. \]