Proof. Let $\ell \in \{1, 2, \dots, n\}$. For a complex number $\alpha \in \mathbb{C}$, consider
$$\begin{aligned} Q(z) &= P(z)(1 + \alpha z^{\ell}) \\ &= \alpha a_n z^{n+\ell} + \dots + \alpha a_{n-\ell+1} z^{n+1} + (a_n + \alpha a_{n-\ell}) z^n + \dots + (a_{\ell} + \alpha a_0) z^{\ell} \\ &\quad + a_{\ell-1} z^{\ell-1} + \dots + a_0. \end{aligned}$$
Take $M > 2n$, then
$$\begin{aligned} \sum_{j=1}^{M} \left| Q \left( e^{i \frac{2j\pi}{M}} \right) \right|^2 &= \sum_{j=1}^{M} \left| P \left( e^{i \frac{2j\pi}{M}} \right) \right|^2 \cdot \left| \left( 1 + \alpha e^{i \frac{2j\pi}{M}\ell} \right) \right|^2 \\ &\le \sum_{j=1}^{M} \left| \left( 1 + \alpha e^{i \frac{2j\pi}{M}\ell} \right) \right|^2 = M(1 + |\alpha|^2). \end{aligned}$$
On the other hand,
$$\sum_{j=1}^{M} \left| Q \left( e^{i \frac{2j\pi}{M}} \right) \right|^2 = M \cdot \left( |\alpha a_n|^2 + \dots + |\alpha a_{n-\ell+1}|^2 + \sum_{j=\ell}^{n} |a_{n-j+\ell} + \alpha a_{n-j}|^2 + |a_{\ell-1}|^2 + \dots + |a_0|^2 \right).$$
Combining the above two estimates, we get
$$|\alpha|^2(|a_n|^2 + \dots + |a_{n-\ell+1}|^2) + \sum_{j=\ell}^{n} |a_{n-j+\ell} + \alpha a_{n-j}|^2 + |a_{\ell-1}|^2 + \dots + |a_0|^2 \le 1 + |\alpha|^2.$$
In particular,
$$|\alpha a_n|^2 + |a_n + \alpha a_{n-\ell}|^2 \le 1 + |\alpha|^2.$$
Choosing $\alpha \in \mathbb{C}$ such that $|\alpha| = \frac{1}{|a_n|}$ and $\arg(a_n) = \arg(\alpha a_{n-\ell})$ in the above equation, we get
$$1 + \left( |a_n| + \frac{|a_{n-\ell}|}{|a_n|} \right)^2 \le 1 + \frac{1}{|a_n|^2}.$$
Solving this gives $|a_{n-\ell}| \le 1 - |a_n|^2$. The proof is complete.